MATH 2441
Probability and Statistics for Biological Sciences
The Binomial Probability Distribution
The binomial probability distribution is the probability distribution of a discrete random variable. The
binomial distribution arises in a generalization of the simple coin-toss experiment we looked at briefly near
the beginning of the course.
The Binomial Experiment
To understand how to apply the binomial distribution in specific problems of interest in biological sciences,
you need to be familiar with the basic characteristics of the so-called binomial experiment. This is a
general description of a common set of features:
1.
2.
3.
4.
The experiment consists of a fixed number, n, of identical trials (or repetitions of some
operation or observation).
Each trial results in one of the same two possible outcomes, which are conventionally denoted
by the characters S (for "success") or F (for "failure").
The successive trials are independent in the sense that the outcome in one trial doesn't
influence the outcome in the next or any of the other trials in the experiment.
The probability of S is the same in each trial, and is conventionally denoted by the letter 'p'.
We denote the complementary probability 1 - p of the outcome F in each trial with the letter 'q'.
While the letters "S" and "F" have traditionally been used to denote the two possible outcomes of each
identical trial, any other pair of characters will do as well. Thus, when we considered the coin flip
experiment, we used the letters H (for "heads") and T (for "tails) to denote the outcome of each flip. In a
way, the designations "success" and "failure" for the two possible outcomes of each trial may be a bit
misleading because of their tendency to indicate "good" or "bad" outcomes in some way. Just think of
"success" as being the outcome of interest, and "failure" as being the other one.
Then, the binomial random variable, x, is defined to be the variable whose value in the number of times S
occurs in the n trials of the binomial experiment. Obviously the possible values of x are whole numbers
ranging between 0 (no S's occurred in the n trials) all the way up n (if all of the trials have S as an outcome).
The binomial probability distribution is the specification of the probability that a specific value of x will be
observed when a binomial experiment is conducted.
To assess whether a random variable is a binomial random variable, or whether the binomial probability
distribution might be useful in solving a probability problem, you simply have to check whether the process
of obtaining the value of the random variable involved meets the four conditions of the binomial experiment.
Example 1: Flipping a fair coin ten times and counting the number of heads obtained is an example of a
binomial experiment. First, the procedure itself consists of n = 10 identical trials, each consisting of a coin
flip. Secondly, each coin flip (or trial) results in one of the same two possible outcomes: H or T. Thirdly,
flipping H in one trial does not influence the outcomes of the other trials in any way. Fourthly, the probability
of getting heads is p = 0.5, the same in each trial.
Example 2: Suppose that it is known that 42% of all flies in a certain population have red eyes. Then,
selecting samples of 20 flies randomly and tallying the number with red eyes is a binomial experiment. First,
we have a fixed number n = 20 of trials, each trial consisting of trapping a fly at random and recording
whether its eye color is red. Secondly, each trial produces one of the same two outcomes: S = fly has red
eyes, or F = fly does not have red eyes. Thirdly, since the flies examined are trapped at random, the trials
are independent -- trapping one particular fly as one of the 20 is unlikely to have any influence on which of
the large number of remaining flies will be selected in other trials. (Of course, you may argue that unless we
release the fly at the end of each trial, those flies already caught cannot be caught again, but this sampling
without replacement option will have negligible consequences if the population of flies is many, many times
larger than 20). Fourthly (again, assuming the population of flies is very large compared to the number 20,
or trapped flies are released and available for trapping another time) the proportion of red-eyed flies will
remain as 42% of the population, and so the probability of trapping a red-eyed fly in any one trial will always
be 0.42.
© David W. Sabo (1999)
The Binomial Distribution
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Example 3: A manufacturer claims that 9.56% of the cans of tomato soup they produce are randomly
underfilled. The experiment of selecting 15 cans of their soup and random and determining how many are
underfilled is a binomial experiment. It consists of n = 15 identical trials, each of which amounts to selecting
a can of tomato soup at random and measuring its contents. Each such trial results in one of the same two
possible outcomes: S = the can is underfilled, or F = the can is not underfilled. Thirdly,the trials are
independent in the sense that having selected one can which is underfilled doesn't make it more or less
likely that the next can to be selected will be underfilled. Finally, the probability of each trial resulting in the
recording of an underfilled can is the constant p = 0.0956.
Example 4: Drawing 20 cards from a deck of 52 cards without replacement and counting the total number
of aces drawn is not a binomial experiment. True, it does consist of a fixed number n = 20 of identical trials,
each involving the selection of one card from the remaining deck of cards. (However, you might question
whether the trials are really identical, since the deck from which each successive card is selected is smaller
than the deck from which the preceding card was selected.) True, each trial has one of the same two
outcomes: S = card is an ace and F = card is not an ace. However, the trials are not independent. If the
first card selected is an ace, then the probability of selecting an ace on subsequent trials is reduced
(because there's fewer aces left for possible selection). In fact, as the experiment proceeds through the 20
trials, the probability, p, of selecting an ace in any given trial will vary up and down as aces or non-aces are
removed one by one from the deck. So, at the very least, properties three and four of the binomial
experiment are violated here, and so this cannot be a binomial experiment. (If instead, after each card was
selected and noted, it was returned to the deck and the deck was reshuffled before the next card was
selected, this would become a binomial experiment -- you should be able to rehearse in your mind why this
modified procedure satisfies all of the conditions of being a binomial experiment.)
Binomial Probabilities
Given the value of n, p, and k, we could use a branching diagram to work out the probability
b(n, p, k) = Pr (x = k, in n trials, with p = probability of S in each trial)
(Bin - 1)
The method is much the same as we used in working out probabilities for simple coin flip experiments with
either fair coins or unfair coins. However, it is possible to analyze the binomial experiment somewhat more
abstractly, so that we can get a direct formula for this probability. You can find the details in various
references on probability. The result is
b(n, p, k )  Cn, k pk qnk
(Bin - 2a)
where
C n, k 
n!
k! n  k !
(Bin - 2b)
(You've seen this numerical coefficient, Cn, r before as the number of combinations of k things that can be
selected from n things. Your calculator may have a function key to calculate these numbers directly, but if
not, you can use the factorial function key which will certainly be on any scientific calculator.)
Example 5: Suppose 8 flies are selected at random from a population in which 42% of all flies have red
eyes. What is the probability that 3 of these flies will have red eyes?
Solution
This looks like a binomial experiment with n = 8 identical trials of selecting a fly and noting whether it has red
eyes, p = 0.42 is the probability of selecting a red-eyed fly in any one trial, and the question is asked for the
probability that we will observe exactly x = 3 red-eyed flies in the 8 trial experiment. Thus, we need to
calculate
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The Binomial Distribution
© David W. Sabo (1999)
b(8, 0.42, 3) 
8!
8!
0.42 3 0.58 5  0.2723
(0.42 ) 3 (1  0.42 ) 8  3 
3! 8  3!
3! 5!
rounded to four decimal places. Thus, in just over 27% of the times this 8-fly experiment is performed,
exactly three of the eight selected flies will have red eyes.

Example 6: It is thought that 25% of all fish caught in a certain vicinity are infected with a particular
parasite. What is the probability that 6 of 20 randomly selected fish from this population will be infected with
the parasite?
Solution
Once again, this example describes a binomial experiment is we can assume that the total number of fish
available for the sampling process is many times greater than 20. Then, this will be a binomial experiment
with n = 20 trials, each consisting of selecting a fish randomly from the population. Each trial will result in
one of the two outcomes: S = the fish is infected with the parasite, or F = the fish is not infected with the
parasite. Since the individual fish are selected randomly, the trials are independent, and the probability of
obtaining an infected fish in any trial is p = 0.25. We are asked to calculate the probability
b(20, 0.25, 6) 
20!
0.25 6 0.75 14  0.1686
6! 14!

Example 7: What is the probability that there will be six or seven infected fish in the sample of twenty fish
selected randomly as described in Example 6 just above?
Solution
The events
A = exactly six of the 20 fish are infected with the parasite
and
B = exactly seven of the 20 fish are infected with the parasite
are mutually exclusive -- if A occurs, then B cannot have occurred, or, if B occurs, then A cannot have
occurred. Then, we recall from our earlier discussion of the basic properties of probabilities that
Pr (A or B ) = Pr(A) + Pr(B)
That is, the probability of A or B occurring is just the sum of their individual probabilities. So, the answer to
the question here is
Pr (six or seven of the 20 fish are infected) = b(20, 0.25, 6) + b(20, 0.25, 7)
We already know that b(20, 0.25, 6)  0.1686 from the result of Example 6. We can calculate that
b(20, 0.25, 7) 
20!
0.25 7 0.75 13  0.1124
7!13!
Thus
Pr (six or seven of the 20 fish are infected)  0.1686 + 0.1124 = 0.2810.
Thus, we can say that approximately 28% of such 20-fish samples will contain six or seven fish which are
infected by this particular parasite.

© David W. Sabo (1999)
The Binomial Distribution
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This last example is a bit more typical of the types of questions asked when the binomial distribution is
applicable: the probability that one of a range of values of the random variable result from the experiment.
One way to calculate such probabilities is to calculate the probability of each of the values in the specified
range individually and sum them, as we did in Example 7. For ranges of more than two or three values, this
can be a rather tedious process. To help expedite this process, it is customary to resort to cumulative
binomial probabilities, for which extensive tables can be constructed for found in published references.
Cumulative Binomial Probabilities and Their Use
Cumulative probabilities can be defined for any probability distribution. We'll describe the concept and
illustrate their properties and use in connection with the binomial distribution.
Basically, a cumulative probability for a random variable x is just a probability of the form Pr (x  k). If x is a
discrete random variable the distinction between  and < in this symbol is important. Thus, a
cumulative probability is just the probability that x has a value of k or less, where k is some possible
outcome of the experiment.
We will use the symbol B(n, p, k) to denote cumulative binomial probabilities, which are formally defined as
B(n, p, k )  Pr( x  k )
k
 Pr( x  0)  Pr( x  1)  Pr( x  2)    Pr( x  k )   Pr( x  j)
j0
k
 b(n, p, 0)  b(n, p,1)  b(n, p, 2)    b(n, p, k )   b(n, p, x)
x 0
(Bin - 3)
This may look rather complicated at first glance, but you just have to remember that B(n, p, k) is simply the
sum of binomial probabilities for the values of x = 0 through x = k (again: the equalities in the  symbols
throughout this document are very important).
One other wrinkle here: there is no simple formula expressing B(n, p, k) in terms of n, p, and k. However,
tables of cumulative binomial probabilities are easily prepared (based in part on just using a computer to do
the sums indicated in (Bin - 3) above). An abbreviated version of such a tabulation is included at the end of
this document. For larger values of n, there are circumstances in which we can use standard normal
probabilities to estimate values of the B(n, p, k) fairly accurately as well.
First, however, we need to demonstrate how to use these cumulative probabilities to solve actual binomial
probability problems. Many textbooks and other references give lists of formulas covering various cases.
However, you don't really need the formulas if you have a good mental picture of what's going on -- or
perhaps, more correctly, you can generate the necessary "formulas" on the fly from an understanding of
what these cumulative probabilities really amount to. Study the following examples.
Example 8: Return to the situation described in Example 6 above, where It is stated that 25% of all fish
caught in a certain vicinity are infected with a particular parasite. What is the probability that between 5 and
10 inclusive of 20 randomly selected fish from this population will be infected with the parasite?
Solution
This clearly still describes a binomial experiment for the reasons discussed in Example 6. Now, we are
asked to find the probability that the experiment will result in either 5 or 6 or 7 or 8 or 9 or 10 infected fish
(the word inclusive means "include the boundary values 5 and 10 of the interval mentioned"). We could
proceed as in Example 7, and simply calculate these six simple probabilities and sum them up to get the
required answer:
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The Binomial Distribution
© David W. Sabo (1999)
Pr(5  x  10) = b(20, 0.25, 5) + b(20, 0.25, 6) + b(20, 0.25, 7) + b(20, 0.25, 8)
+ b(20, 0.25, 9) + b(20, 0.25, 10)
but this is beginning to look like quite a lot of work. Unfortunately, the sum of probabilities here isn't
precisely a cumulative probability either, because it doesn't include terms with x, the third index, going all the
way down to zero. But, note the following:
wanted:
sum over
outcomes
B(20, 0.25, 4) =
sum over
outcomes
0
1
2
3
4
B(20, 0.25, 10) =
sum over
outcomes
0
1
2
3
4
5
6
7
8
9
10
5
6
7
8
9
10
Thus, it looks like the following is true:
B(20, 0.25, 10) = B(20, 0.25, 4) + Pr(5  x  10)
and so, we can calculate the desired answer to the present problem using the formula
Pr(5  x  10) = B(20, 0.25, 10) - B(20, 0.25, 4)
the difference of two cumulative probabilities. You can see from this that at its most complicated, any
binomial probability can be calculated as the difference of two cumulative probabilities. Sometimes
what a question asks for amounts to just a cumulative probability directly, and so even this subtraction is
unnecessary.
If you refer to the cumulative binomial probability table at the end of this document, you'll notice that it is
arranged in sections, one for each value of n represented. Here, n = 20, so you'll need to go to the middle
part of the third page of our tables, to the section labeled n = 20 at the extreme left. Then, you'll notice that
the columns of the table correspond to various values of p. Here we have p = 0.25, so we will need to use
the fourth column headed by that value. Finally, we need cumulative probabilities for x  10 and for x  4, so
we read from the rows in the n = 20 section that are labeled x = 10 and x = 4, respectively, getting
B(20, 0.25, 10)  0.9961
B(20, 0.25, 4)  0.4148
(Values in these tables are rounded off to four decimal places, which is adequate for most applications.)
Thus, the answer to the question asked above is simply
Pr(5  x  10) = B(20, 0.25, 10) - B(20, 0.25, 4)  0.9961 - 0.4148 = 0.5813
Thus, there is a probability of 0.5813 that the number of infected fish in the sample of 20 will be a number
betrween 5 and 10 inclusive.

Example 9: What is the probability that more than ten of the twenty fish described in Example 6 will have
parasites? What would be this probability if, instead of being told that 25% of the population of fish were
infected with parasites, you were told that 60% of the fish were infected with parasites?
Solution
We have been asked to calculated Pr(x > 10). Note that this is not the same thing as Pr(x  10) -- with
discrete probability distributions such as the binomial, there may well be a significant difference between the
"greater than" and "greater than or equal" events.
Although the cumulative probabilities are particularly suited to calculating probabilities of "less than" type
events, it isn't too hard to get a "greater than" type probability out of them. Note the following:
© David W. Sabo (1999)
The Binomial Distribution
Page 5 of 19
B(20, 0.25, 20) 
sum over
outcomes
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
B(20, 0.25,10) 
sum over
outcomes
0 1 2 3 4 5 6 7 8 9 10
x > 10 
sum over
outcomes
11 12 13 14 15 16 17 18 19 20
From this diagram, it looks like we can write
P(x > 10) = Pr(0  x  20) - Pr(0  x  10)
= B(20, 0.25, 20) - B(20, 0.25, 10)
= 1 - 0.9961 = 0.0039
The tables do not contain an explicit value for B(20, 0.25, 20) = Pr(0  x  20). But since this probability is
the sum of the probabilities of all of the possible values that x can attain in this n = 20 experiment, we know
that cumulative probabilities for k = n must always be equal to 1.
Anyway, there is the relatively small probability of 0.0039 (or about 1 in 250) that if 25% of the entire
population of fish is infected with the parasite, a random sample of 20 fish will have more than 10 infected
members.
On the other hand, if we were informed that 60% of the population of fish was infected with the parasite,
then the probability of observing more than 10 infected fish in a random sample of 20 would be
P(x > 10) = Pr(0  x  20) - Pr(0  x  10)
= B(20, 0.60, 20) - B(20, 0.60, 10) = 1 - 0.2447 = 0.7553
When 60% of the population of fish carries the parasite, there is a probability of 75.53% that a random
sample of 20 fish will have more than 10 infected members.

Example 9a: Suppose you were told that 25% of the population of these fish were infected with the
parasite, and you examined a random sample of 20 of the fish and found eleven that were infected. What
does this observation make you think about the original report that only 25% of the population is infected,
and why?
Solution
The calculations in the solution to Example 9 above indicate that if only 25% of the population of fish is
infected with the parasite, then the observation of more than ten infected fish is expected to occur in only
about 1 in 250 samples of 20 fish from that population. However, you would have just observed one of
those extremely rare events. You will recall from our discussion of statistical hypothesis testing that we
consider very unexpected outcomes to be evidence that our expectation is suspect, rather than that we've
just observed a very unexpected coincidence. In the same way here, most practitioners would suspect that
having observed this outcome of more than 10 infected fish in the random sample of 20, there is reason to
believe that such a high rate of infection isn't as improbable as the we thought -- that is, rather than simply
dismiss this observation as a rare coincidence, we suspect that it isn't all that rare. Working backwards
then, we would suspect that the rate of infection in the fish population could well be much higher than 25%.
How much higher? Well, the 25% quoted would be a proposed estimate of the population proportion, , of
infected fish. We would need to compute a confidence interval estimate for this proportion. Technically, the
data presented here is enough to validate the large sample confidence interval estimation formulas for a
population proportion according to the usual rule-of-thumb. The random sample of 20 fish included 11 that
were infected (and 11 > 5) and so 9 that were not infected (and 9 > 5). When we calculate, say, a 95%
confidence interval estimate of the population proportion with this data, we get
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The Binomial Distribution
© David W. Sabo (1999)
  p  z 0.025

p 1  p
n
11
 1.96 
20
1120 1  1120 
20
= 0.55  0.22 @95%
or
0.33    0.77 @95%
Thus, while we have the data to validate the large sample estimation process here, the sample size is still so
small that the resulting 95% confidence interval estimate is rather imprecise, having an uncertainty almost
as large as the 25% figure originally quoted as the population proportion. Nevertheless, at a 95%
confidence level, we can say that the true population proportion is definitely larger than the 0.25 quoted in
the original example.

Example 10: Hank believes that if a coin is fair, then the probability it will fall heads is the same as the
probability it will fall tails, and so both are exactly 0.5. Thus, when he participates in a little game of chance
with Clem, with Clem flipping one of his own coins ten times and getting 6 heads and 4 tails, Hank accuses
Clem of cheating. Hank states that if Clem had not been cheating, then the experiment would have
undoubtedly ended up in 5 heads and 5 tails. Does Hank have a case?
Solution
Well, there's only one way to decide. We need to determine the probability that that a binomial experiment
with n = 10 trials and a probability of S = Heads, say, on each trial of exactly 0.5, will result in exactly five S's
and 5 F's (where if S = Heads, then obviously F = Tails). Thus, if x is the number of heads that turns up in
this experiment, we are asked to find
Pr(x = 5 exactly, n = 10, p = 0.5)  b(10, 0.5, 5)
Now, we have a formula for this probability -- formulas (Bin - 2a, b) :
b10, 0.5, 5 
10!
0.55 0.55  0.2461
5! 5!
rounded to four decimal places. Also, noticing that
B(10, 0.5, 5) =
sum over
outcomes
0
1
2
3
4
B(10, 0.5, 4) =
sum over
outcomes
0
1
2
3
4
difference =
b(10, 0.5, 5)
5
5
we see that we can calculate individual binomial probabilities as the difference of two cumulative binomial
probabilities:
b(10, 0.5, 5) = B(10, 0.5, 5) - B(10, 0.5, 4)
= 0.6230 - 0.3770 = 0.2460
(The discrepancy of 1 unit in the fourth decimal place between this result and the one obtained just above
using the algebraic formula for b(10, 0.5, 5) is due to rounding off to four decimal places in the table entries.
© David W. Sabo (1999)
The Binomial Distribution
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When we round to four decimal places, we expect that arithmetic with such numbers may be inaccurate by
up to a few units in the fourth decimal place. When we calculate the cumulative probabilities more
accurately, say to 7 decimal places, we get B(10, 0.5, 5) = 0.6230469 and B(10, 0.5, 4) = 0.3769531, so that
the difference becomes 0.2460938, which agrees with the result from formulas (Bin - 2a, b) now when
rounded to seven decimal places. You needn 't worry about probabilities that are out by 1 unit in the fourth
significant figure -- such an imprecision has no practical significance.)
From this calculation, you see that even with a perfectly fair coin, we would only expect to get exactly five
heads in ten flips about one quarter of the times that the 10-flip experiment is performed -- not a rare
occurrence, but far from a certain occurrence. When we say that a fair coin will fall heads and tails an equal
number of times over the long run, we mean that over a very large number of coin flips (for absolute
exactness, this would really have to be an unimaginably large number), the fraction of flips that result in
heads would become nearer and nearer to 0.5. However, for a relatively small number of coin flips, such as
10, getting exactly 50% heads is less likely. (In fact, if you flipped the fair coin an odd number of times, it
would obviously be impossible to get heads exactly 50% of the time, because that would require you to get a
number of heads which is not a whole number!)
Probability
We can illustrate this effect with the diagram to the
Comparison of Binomial Probabilities
right. The smooth line curving downwards to the
1
right represents the probability of getting exactly
0.9
50% heads when a fair coin is tossed n times, with n
0.8
ranging from 2 to 300 as indicated along the
0.7
horizontal axis. The more jagged "curve" arcing up
0.6
towards the right is the probability of getting
0.5
between 45% and 55% heads in n tosses. (The
0.4
jaggedness results because we need to round 0.45n
0.3
and 0.55n off to whole numbers to calculate the
0.2
probabilities. What we really did to prepare the
0.1
figure was to calculate the interval as 0.5n  0.05n,
0
0
50
100
150
200
250
300
but with 0.05n rounded off to an integer. Thus, for n
Number of Trials (n)
= 28, 0.05n = 1.4 and so rounds to 1, but for n = 30,
0.05n = 1.5 and so rounds to 2. When n = 28, the
interval of x values used in the graph is 14  1, or x = 13, 14 and 15. But for n = 30, the interval of x values
used in the graph becomes 15  2 or the five values x = 13, 14, 15, 16, and 17. This increase from including
three outcomes to including five outcomes causes the sudden jump in the graph at n = 30. As n gets larger,
this effect becomes less significant. For the present purposes, you just need to consider the trend of the
jagged graph to rise approaching closer and closer to the value 1.)
This diagram demonstrates that while the probability of getting exactly 50% heads in n tosses of a fair coin is
actually dropping off effectively to zero as n increases arbitrarily, the probability of getting a value in the
close vicinity of 50% of the tosses is becoming closer and closer to 1. As the number of tosses increases, it
becomes more and more likely that you will observe a number of heads which is close to 50%, but very
unlikely that you'll ever see exactly 50% of the tosses resulting in heads.

Example 11: Due to a variety of circumstances he believes to have been beyond his control, Hank arrives
at a statistics midterm test utterly unprepared. (In fact, he arrived a few minutes late because his schedule
hadn't really permitted him to attend class conveniently so far, and so it took a while to find the room!).
Anyway, the test consisted of just twelve questions and problems, with five multiple choice possible
responses in each case. If Hank's ignorance is total, so that he must simply select one of the five possible
answers for each question by sheer random guess, what are his chances of passing the test? What if Hank
is able to eliminate two possible answers in each question as clearly wrong, so that he needs to guess
between just 3 possible answers in each case? Assume a pass mark is 50% or six answers correct.
Solution
Start by reviewing the details of this little probabilistic "experiment" by Hank. First, Hank will be guessing the
answer to each of the 12 questions, one at a time. He is performing 12 identical "trials." Each of these trials
will have one of two possible outcomes: either he guesses the correct answer (probability = 1/5 = 0.2 in the
first instance where he is unable to eliminate any of the possible answers in each question before guessing)
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The Binomial Distribution
© David W. Sabo (1999)
or he guesses a wrong answer. The trials are independent, since whatever method he uses to decide which
answer to guess, we can assume that the instructor has been devious enough to place the actual correct
answers in random positions as you move from one question to the next. Thus, guessing the correct answer
for one question will not affect the likelihood of guessing the correct answer on any subsequent question.
The probability of guessing the correct answer will continue to be 0.2 for each question.
So, although Hank probably doesn't know it, he is really performing a binomial experiment here! There are n
= 12 trials, each with the same two possible outcomes (S = correct answer chosen, F = correct answer not
chosen). If we let x stand for the number of correct answers chosen by Hank, then the question is asking us
to find Pr(x  6), the probability he will answer six or more of the questions correctly.
To be able to use the cumulative binomial probability tables, we first need to express Pr(x  6) as a "" type
probability. This is easily done, as illustrated in the following diagram:
all possible outcomes
includes x =
0
1
2
3
4
5
event x  5
includes x =
0
1
2
3
4
5
event x  6
includes x =
6
7
8
9
10
11
12
6
7
8
9
10
11
12
Thus, since the events "x  5" and "x  6" together cover all the possible outcomes in this situation, we can
write
Pr(x  6) = 1 - Pr(x  5)
But Pr(x  5) has the form of a cumulative probability, and so
Pr(x  5) = B(12, 0.2, 5)  0.9806
from the table of cumulative binomial probabilities. Thus,
Pr(x  6) = 1 - Pr(x  5)  1 - 0.9806 = 0.0194.
Thus, based on total random guesswork, Hank has a probability of 0.0194 (or about 2%) of passing this test
-- a bit of a long shot.
Now, if Hank can reliably eliminate two of the five possible answers for each question, so that he is guessing
between only three possibilities in each case, his probability of correctly guessing the answer to any one
question rises to 1/3 or 0.333 to three decimal places. Then, his probability of passing the test becomes
Pr(x  6) = 1 - Pr(x  5) = 1 - B(12, 0.333, 5)
Our cumulative binomial probability tables do not have entries for p = 1/3 (the nearest is for p = 0.3 and
p = 0.4). We can do one of several things here.
First, be conservative: use the p = 0.3 value as the next lowest value of p in the table. This would give
Pr(x  6) = 1 - Pr(x  5)  1 - B(12, 0.3, 5)  1 - 0.8822 = 0.1178.
Secondly, we could use linear interpolation between the table entries for p = 0.3 and p = 0.4. Since
B(12, 0.3, 5) = 0.8822
and
B(12, 0.4, 5) = 0.6652
we could estimate that
B(12, 0.333, 5)  B(12, 0.3, 5) + [B(12, 0.4, 5) - B(12, 0.3, 5)]/3
= 0.8822 + [0.6652 - 0.8822]/3  0.8099
© David W. Sabo (1999)
The Binomial Distribution
Page 9 of 19
This would give
Pr(x  6) = 1 - Pr(x  5) = 1 - B(12, 0.333, 5)  1-0.8099 = 0.1901
Finally, we could use a computer application that will give us exact values of B(n, p, x) for any values of p.
For example, the Excel function BINOMDIST(5, 12, 0.333,TRUE) gives the result 0.8229 to four decimal
places for what we've been calling B(12, 0.333, 5). This gives
Pr(x  6) = 1 - Pr(x  5) = 1 - B(12, 0.333, 5)  1-0.8229 = 0.1771
This last value, 0.1771, is essentially exact through four significant figures, giving you an indication of the
accuracy of the linear interpolation process used just above.
At any rate, it looks like that by being able to reliably eliminate two of the five possible answers for each
question, Hank has increased his chances of passing the test from about 2% to about 18%, an increase in
likelihood by a factor of 10. His odds are still not very good for passing the test, but this example does
demonstrate the benefit of being able to eliminate from consideration answers which are clearly wrong when
you are answering questions for which you aren't sure of the correct answer on a multiple choice test.

We will present just one more example here, which is a simple illustration of one of the more important
applications of the binomial distribution. This example is an illustration of what is often called a "lotacceptance" problem -- an experiment in which the decision to accept or reject an entire "lot" or shipment
of some items is based on the detailed examination of just a small random sample of items from the total.
This example gives you a bit of a hint of what might be involved in testing hypotheses about population
proportions based on small samples. As you work through this example, you will see familiar concepts from
hypothesis testing appearing.
Example 12 A food processing company buys potatoes by the truckload from a supplier. They have certain
specifications involving size, shape, texture, and so forth they wish the potatoes to satisfy. In fact, their
target is to accept only those truckloads of potatoes for which they believe 90% of the potatoes satisfy their
specifications, and so to try as much as possible to reject truckloads of potatoes for which more than 10% of
the potatoes fail to satisfy their specifications. Since it would be impractical to examine every single potato
on a truckload, they devise the following decision rule: they will select 25 potatoes at random from the
truckload. If 21 or more meet their specifications, they will accept the entire truckload of potatoes. However,
if 20 or fewer meet their specifications, they will reject the entire truckload, returning it to the supplier. To
evaluate the effectiveness or appropriateness of this decision rule, calculate the probability that they will
reject a truckload in which 90% of the potatoes overall do meet their specifications, and also, calculate the
probability that they will accept a truckload in which only 80% of the potatoes meet their specifications.
Solution
(You should already see that this question is really testing the hypotheses "the proportion of potatoes in the
truckload meeting specifications is 90% or more" vs. "the proportion of potatoes in the truckload meeting
specifications is less than 90%." There are two possible types of errors here: they can reject a truckload in
which 90% or more of the potatoes actually satisfy their specifications or they can accept a truckload in
which less than 90% of the potatoes meet their specifications. These errors can arise, of course, because
the company is basing its assessment of the entire truckload on a random sample of just 25 potatoes. The
decision rule described in the example, is, of course, simply an instance of what we would have called the
rejection criterion in hypothesis testing.)
It is clear that this problem involves a binomial experiment. A fixed number of identical trials are performed,
each consisting of selecting a potato at random for examination. Each trial results in one of the same two
outcomes: the potato either satisfies the specifications or it does not. Since the potatoes are randomly
selected from the entire shipment in some fashion, the trials are independent. Finally, the proportion of good
and bad potatoes in the truckload will not be affected in any perceptible way by the selection of 25 potatoes,
and so we can say that the probability of getting a good potato or of getting a bad potato in each of the trials
is the same throughout the experiment, and equal to the corresponding proportion of potatoes in the
truckload as a whole.
Page 10 of 19
The Binomial Distribution
© David W. Sabo (1999)
Thus,
Pr (a truckload will be rejected even if 90% of the potatoes satisfy specifications)
= Pr(fewer than 21 of 25 potatoes satisfy specifications when 90% of the truckload does)
= Pr(x  20 in n = 25 trials with p = 0.90)
Here, we are using x to stand for the number of potatoes in the sample of 25 which satisfy the specifications.
This last expression is already in the form of a cumulative probability, so we can write immediately
Pr (a truckload will be rejected even if 90% of the potatoes satisfy specifications)
= B(25, 0.9, 20)
= 0.0980
Thus, there is just under a 10% chance that the company will reject a truckload of potatoes that really just
barely satisfies their desired specifications. On the other hand
Pr(a truckload will be accepted even if only 80% of the potatoes satisfy specifications)
= Pr(21 or more of the 25 potatoes sampled satisfy specs when 80% of the total do)
= Pr(x  21 when n = 25 and p = 0.80)
Here, x is again representing the number of potatoes in the sample that satisfy the company's specifications.
The next step is to turn this probability into the form of a cumulative probability. Note that the event "21 or
more" has the complement "20 or fewer."
Pr(a truckload will be accepted even if only 80% of the potatoes satisfy specifications)
= 1 - Pr(x  20 when n = 25 and p = 0.80)
= 1 - B(25, 0.80, 20)
= 1 - 0.5793
= 0.4207
Thus, there is a probability of over 40% that in using this decision rule, the company will accept a truckload
of potatoes in which only 80% of the potatoes satisfy their specifications.
Neither of the error probabilities calculated here are particularly small, but the second one being not that far
away from 50% (which you could achieve by simply flipping a coin!) seems to be a cause for concern. It is
clear that the decision rule described in this example is not very reliable overall. However, how to fix it up is
a problem that we don't really have time to address at greater length in this course except to make several
general observations:
(i)
The probability of accepting a bad truckload of potatoes drops off sharply with the as the proportion
of potatoes satisfying the specifications continues to decrease. We found that there was a
probability of 0.4207 of accepting a truckload in which only 80% of the potatoes satisfied
specifications. By reading along the x = 20 row of the n = 25 cumulative binomial probability table,
you can see that that probability drops to 0.2137 for a 70% truckload, 0.0905 for a 60% truckload,
0.0095 for a 50% truckload, and so on. The farther a truckload of potatoes is from the target
proportion satisfying specifications, the more likely the rule is to result in a "reject" decision. If our
tables were more extensive, you'd also be able confirm that as the proportion of potatoes on the
truckload which satisfy specifications increases from 90% towards 100%, the probability of rejecting
the truckload also decreases sharply.
(ii)
We can't really improve the overall situation by keeping the number of sampled potatoes at 25 and
adjusting the cutoff value from, say, 21 to 22 (that is, by changing the rule to read "reject the entire
© David W. Sabo (1999)
The Binomial Distribution
Page 11 of 19
truckload unless 22 or more of the 25 potatoes satisfy specifications). You can easily verify that
this does indeed reduce the probability of accepting a truckload in which only 80% of the potatoes
satisfy specifications from 0.4207 to 0.2340 (which is still not great, but better), but at the same
time, now the probability of rejecting a truckload in which 90% of the potatoes satisfy specifications
will rise to 0.2364 (that these two probabilities are nearly the same is coincidence here). This is
exactly the problem we found in hypothesis testing in general: if you keep everything about the
experiment the same -- particularly sample sizes -- and simply move the boundary of the rejection
region one way or the other, you will reduce the probability of making one type of error, but only at
the cost of increasing the probability of making the opposite error.
(iii)
Finally, the last sentence suggests the only really effective way to achieve a decision rule in which
the probabilities of both types of error described in this problem can be controlled to be smaller
than some acceptable target values. You will need to base the decision on a sample which is
larger than 25 potatoes. Procedures have been developed to assist the design of such decision
procedure. Most references on sampling theory or quality control should provide details.

The Normal Approximation to the Binomial Distribution
The biggest problem encountered in working with the binomial probability distribution is that the tables of
cumulative probabilities either cover only a limited range of experiments (that is only certain smaller values
of n) and only certain values of p. Compared with the single-page table that we were able to use to
calculate normal probabilities to adequate accuracy for virtually any practical situation, we found in Example
10 above that it is quite easy to come up with problems that aren't really adequately covered by our four
pages of cumulative binomial probabilities
Of course, with the easy access to powerful computer applications these days, we tend to rely less and less
on printed tables of limited scope for calculating probabilities, and instead can make use of computer
procedures which are capable of quickly and accurately determining the specific probabilities we require.
However, there is one approximation formula for binomial probabilities that may still be useful in its own right
as a quick way to compute approximate cumulative binomial probabilities, and also sheds light on the
formulas we earlier stated without proof in connection with estimation and testing hypotheses involving
population proportions. (The connection here is that the probability, p, of S in a single trial of a binomial
experiment can often be interpreted as a population proportion.)
Early in the course, we defined the concepts of mean values and variances of a random variable. For a
discrete random variable x, the formulas are
 x   x k Pr( x  x k )
k
and
 2   x k   x  Pr( x  x k )
2
k
where the summations are carried out over all possible values, xk, of the random variable. Without going
into the details here, it is possible to carry out these summations in general for the binomial random variable,
making use of the properties of formulas (Bin -2a,b), and the result is
and
so that
 x  np
(Bin - 4a)
 2  npq  np1  p
(Bin - 4b)
  npq  np 1  p 
(Bin - 4c)
Note that the mean value of x is just the product of the number of trials and the probability of "success" for
each trial. Thus, the quantity p really can be interpreted as the proportion of trials which result in "success."
Page 12 of 19
The Binomial Distribution
© David W. Sabo (1999)
If we tabulate binomial probabilities, b(n, p, x) for
a specific value of x, and then construct a
histogram of the probabilities so that bars of width
1 and height b(n, p, k) are centered above the
value x = xk, we get a chart that looks very much
like a somewhat jagged bell curve.
0.2
0.15
The diagram to the right illustrates this for n = 20
0.1
and p = 0.5. Here we've joined the centers of the
tops of the bars by straight line segments to
emphasize how bell-like the resulting histogram
really is. This suggests that we might be able to
0.05
use the standard normal probability tables to
calculate approximate values for binomial
probabilities, since the area under the rough bell
0
curve will be approximately equal to the area in
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
the corresponding bars of the histogram (which is
identical to the binomial probability). In fact, you
can see that there will be a tendency for errors in the approximation to cancel each other out, because the
rough bell curve includes small wedges outside the histogram, but also excludes roughly equal-size small
wedges inside the histogram.
From the illustration, you can see that the procedure must be as follows. Suppose x is a binomial random
variable corresponding to an experiment with n trials and p the probability of S in each trial. Then
Pr(a  x  b) = area in the histogram bars corresponding to x = a, x = a + 1, etc. up to x = b.
(Here 'a' and 'b' are, of course, whole numbers.)
The corresponding normal distribution approximating the distribution of x will be a normal distribution with
mean value and standard deviation given by formulas (Bin - 4a, b, c). Note, however that the histogram
bars are centered on the values x = a, x = a + 1, etc. Thus, the area in the histogram bars corresponding to
x = a through x = b is approximately equal to the area under the normal distribution curve for the interval
from a - 0.5 to b + 0.5.
To make the approximation procedure very explicit, we define our symbols now as:
x = a binomial random variable corresponding to an experiment with n trials, and the probability of
outcome "success" in each trial is p.
and
y = a normally distributed random variable with mean np and variance np(1 - p).
Then
Pr(a  x  b)  Pr(a - 0.5  y  b + 0.5)
(Bin - 5)
where the right-hand side is just a normal probability problem, with its value computed using the standard
normal probability tables in the usual way. In fact, at the risk of making things look more complicated than
they really are, we can express (Bin - 5) in terms of the standard normal random variable, z, explicitly:
 a  np  0.5
b  np  0.5 
Pr a  x  b   Pr a  0.5  y  b  0.5   Pr 
z
 np 1  p 
np 1  p  

(Bin - 6)
The subtraction and addition of 0.5 in the normal probability part of formulas (Bin - 5) and (Bin - 6) is
sometimes called the continuity correction. It is nothing more than a simple accounting for the way in
which we construct histograms. This normal approximation to the binomial distribution becomes more
accurate as the value of n increases, and is generally more accurate for values of p near 0.5 than for values
of p near the extremes of 0 and 1. When p is not 0.5, the binomial distribution is not symmetric about its
© David W. Sabo (1999)
The Binomial Distribution
Page 13 of 19
mean value, though as n gets larger, the asymmetry in areas under the histogram becomes less significant.
Thus, as a general rule of thumb, this approximation is recommended only if both np  5 and n(1 - p)  5.
You should recognize these two conditions as the rules of thumb we used for recognizing the large sample
case when doing statistical inference for the population proportion -- this is not a coincidence.
Example 13: Accepting that the frequency of a certain genetic mutation in a population is one in 35,
compute the probability that there will be between 10 and 20 people inclusive with this mutation in a random
sample of 500 individuals selected from that population.
Solution
Under the conditions of random sampling from a population which is much larger than 500, we can satisfy
ourselves that this situation is a binomial experiment, with each individual sampled being classified as either
or
S = has the genetic mutation
F = does not have the genetic mutation
So, letting
x = the number of individuals in the sample with the mutation
we find that we are being asked to determine
Pr(10  x  20) = B(500, 1/35, 20) - B(500, 1/35, 9)
We really have two difficulties here. First, we would need an n = 500 cumulative binomial probability table,
and, secondly, it would have to have a column for p = 1/35  0.02857. Even if we could find a table for n =
500, it is unlikely that it would also include cumulative probabilities for a value of p relatively near to 0.02857.
However, noting that np  14.29 > 5 here, we can use the normal approximation in formula (Bin - 6). We
note that
 1 
 x  np  500     14 .2857
 35 
and
  np 1  p  
500 
1   34 
    3.7253
35
   35 
Thus, with a = 10 and b = 20 in formula (Bin - 6), we get
Pr(10  x  20 )  Pr 10  0.5  y  20  0.5 
20  0.5  14 .2857 
 10  0.5  14 .2857
 Pr 
z

3
.
7253
3.7253


 Pr(  1.28  z  1.67 )
 0.4525  0.3997  0.8522
(What is interesting is that employing the same procedure as was used to set up the cumulative binomial
probability tables distributed in class, we can compute the answer to this problem exactly, and it turns out to
be 0.8528 to four decimal places. The normal approximation is extremely accurate here! Excel's
BINOMDIST function could also have been used to compute the exact solution to this problem.)

Page 14 of 19
The Binomial Distribution
© David W. Sabo (1999)
When the condition np > 5 is not met, cumulative binomial probabilities can be approximated using the
Poisson distribution. Details are available in most standard probability and statistics textbooks.
© David W. Sabo (1999)
The Binomial Distribution
Page 15 of 19
Cumulative Binomial Probabilities
p=
n= x=
0.05
0.1
0.2
0.25
0.3
0.4
0.5
0.6
0.7
0.75
0.8
0.9
0.95
2
0
1
2
0.9025
0.9975
1.0000
0.8100
0.9900
1.0000
0.6400
0.9600
1.0000
0.5625
0.9375
1.0000
0.4900
0.9100
1.0000
0.3600
0.8400
1.0000
0.2500
0.7500
1.0000
0.1600
0.6400
1.0000
0.0900
0.5100
1.0000
0.0625
0.4375
1.0000
0.0400
0.3600
1.0000
0.0100
0.1900
1.0000
0.0025
0.0975
1.0000
3
0
1
2
3
0.8574
0.9928
0.9999
1.0000
0.7290
0.9720
0.9990
1.0000
0.5120
0.8960
0.9920
1.0000
0.4219
0.8438
0.9844
1.0000
0.3430
0.7840
0.9730
1.0000
0.2160
0.6480
0.9360
1.0000
0.1250
0.5000
0.8750
1.0000
0.0640
0.3520
0.7840
1.0000
0.0270
0.2160
0.6570
1.0000
0.0156
0.1563
0.5781
1.0000
0.0080
0.1040
0.4880
1.0000
0.0010
0.0280
0.2710
1.0000
0.0001
0.0073
0.1426
1.0000
4
0
1
2
3
0.8145
0.9860
0.9995
1.0000
0.6561
0.9477
0.9963
0.9999
0.4096
0.8192
0.9728
0.9984
0.3164
0.7383
0.9492
0.9961
0.2401
0.6517
0.9163
0.9919
0.1296
0.4752
0.8208
0.9744
0.0625
0.3125
0.6875
0.9375
0.0256
0.1792
0.5248
0.8704
0.0081
0.0837
0.3483
0.7599
0.0039
0.0508
0.2617
0.6836
0.0016
0.0272
0.1808
0.5904
0.0001
0.0037
0.0523
0.3439
0.0000
0.0005
0.0140
0.1855
5
0
1
2
3
4
0.7738
0.9774
0.9988
1.0000
1.0000
0.5905
0.9185
0.9914
0.9995
1.0000
0.3277
0.7373
0.9421
0.9933
0.9997
0.2373
0.6328
0.8965
0.9844
0.9990
0.1681
0.5282
0.8369
0.9692
0.9976
0.0778
0.3370
0.6826
0.9130
0.9898
0.0313
0.1875
0.5000
0.8125
0.9688
0.0102
0.0870
0.3174
0.6630
0.9222
0.0024
0.0308
0.1631
0.4718
0.8319
0.0010
0.0156
0.1035
0.3672
0.7627
0.0003
0.0067
0.0579
0.2627
0.6723
0.0000
0.0005
0.0086
0.0815
0.4095
0.0000
0.0000
0.0012
0.0226
0.2262
6
0
1
2
3
4
5
0.7351
0.9672
0.9978
0.9999
1.0000
1.0000
0.5314
0.8857
0.9842
0.9987
0.9999
1.0000
0.2621
0.6554
0.9011
0.9830
0.9984
0.9999
0.1780
0.5339
0.8306
0.9624
0.9954
0.9998
0.1176
0.4202
0.7443
0.9295
0.9891
0.9993
0.0467
0.2333
0.5443
0.8208
0.9590
0.9959
0.0156
0.1094
0.3438
0.6563
0.8906
0.9844
0.0041
0.0410
0.1792
0.4557
0.7667
0.9533
0.0007
0.0109
0.0705
0.2557
0.5798
0.8824
0.0002
0.0046
0.0376
0.1694
0.4661
0.8220
0.0001
0.0016
0.0170
0.0989
0.3446
0.7379
0.0000
0.0001
0.0013
0.0159
0.1143
0.4686
0.0000
0.0000
0.0001
0.0022
0.0328
0.2649
7
0
1
2
3
4
5
6
0.6983
0.9556
0.9962
0.9998
1.0000
1.0000
1.0000
0.4783
0.8503
0.9743
0.9973
0.9998
1.0000
1.0000
0.2097
0.5767
0.8520
0.9667
0.9953
0.9996
1.0000
0.1335
0.4449
0.7564
0.9294
0.9871
0.9987
0.9999
0.0824
0.3294
0.6471
0.8740
0.9712
0.9962
0.9998
0.0280
0.1586
0.4199
0.7102
0.9037
0.9812
0.9984
0.0078
0.0625
0.2266
0.5000
0.7734
0.9375
0.9922
0.0016
0.0188
0.0963
0.2898
0.5801
0.8414
0.9720
0.0002
0.0038
0.0288
0.1260
0.3529
0.6706
0.9176
0.0001
0.0013
0.0129
0.0706
0.2436
0.5551
0.8665
0.0000
0.0004
0.0047
0.0333
0.1480
0.4233
0.7903
0.0000
0.0000
0.0002
0.0027
0.0257
0.1497
0.5217
0.0000
0.0000
0.0000
0.0002
0.0038
0.0444
0.3017
8
0
1
2
3
4
5
6
7
0.6634
0.9428
0.9942
0.9996
1.0000
1.0000
1.0000
1.0000
0.4305
0.8131
0.9619
0.9950
0.9996
1.0000
1.0000
1.0000
0.1678
0.5033
0.7969
0.9437
0.9896
0.9988
0.9999
1.0000
0.1001
0.3671
0.6785
0.8862
0.9727
0.9958
0.9996
1.0000
0.0576
0.2553
0.5518
0.8059
0.9420
0.9887
0.9987
0.9999
0.0168
0.1064
0.3154
0.5941
0.8263
0.9502
0.9915
0.9993
0.0039
0.0352
0.1445
0.3633
0.6367
0.8555
0.9648
0.9961
0.0007
0.0085
0.0498
0.1737
0.4059
0.6846
0.8936
0.9832
0.0001
0.0013
0.0113
0.0580
0.1941
0.4482
0.7447
0.9424
0.0000
0.0004
0.0042
0.0273
0.1138
0.3215
0.6329
0.8999
0.0000
0.0001
0.0012
0.0104
0.0563
0.2031
0.4967
0.8322
0.0000
0.0000
0.0000
0.0004
0.0050
0.0381
0.1869
0.5695
0.0000
0.0000
0.0000
0.0000
0.0004
0.0058
0.0572
0.3366
9
0
1
2
3
4
5
6
7
8
0.6302
0.9288
0.9916
0.9994
1.0000
1.0000
1.0000
1.0000
1.0000
0.3874
0.7748
0.9470
0.9917
0.9991
0.9999
1.0000
1.0000
1.0000
0.1342
0.4362
0.7382
0.9144
0.9804
0.9969
0.9997
1.0000
1.0000
0.0751
0.3003
0.6007
0.8343
0.9511
0.9900
0.9987
0.9999
1.0000
0.0404
0.1960
0.4628
0.7297
0.9012
0.9747
0.9957
0.9996
1.0000
0.0101
0.0705
0.2318
0.4826
0.7334
0.9006
0.9750
0.9962
0.9997
0.0020
0.0195
0.0898
0.2539
0.5000
0.7461
0.9102
0.9805
0.9980
0.0003
0.0038
0.0250
0.0994
0.2666
0.5174
0.7682
0.9295
0.9899
0.0000
0.0004
0.0043
0.0253
0.0988
0.2703
0.5372
0.8040
0.9596
0.0000
0.0001
0.0013
0.0100
0.0489
0.1657
0.3993
0.6997
0.9249
0.0000
0.0000
0.0003
0.0031
0.0196
0.0856
0.2618
0.5638
0.8658
0.0000
0.0000
0.0000
0.0001
0.0009
0.0083
0.0530
0.2252
0.6126
0.0000
0.0000
0.0000
0.0000
0.0000
0.0006
0.0084
0.0712
0.3698
10
0
1
2
3
4
5
6
7
8
9
0.5987
0.9139
0.9885
0.9990
0.9999
1.0000
1.0000
1.0000
1.0000
1.0000
0.3487
0.7361
0.9298
0.9872
0.9984
0.9999
1.0000
1.0000
1.0000
1.0000
0.1074
0.3758
0.6778
0.8791
0.9672
0.9936
0.9991
0.9999
1.0000
1.0000
0.0563
0.2440
0.5256
0.7759
0.9219
0.9803
0.9965
0.9996
1.0000
1.0000
0.0282
0.1493
0.3828
0.6496
0.8497
0.9527
0.9894
0.9984
0.9999
1.0000
0.0060
0.0464
0.1673
0.3823
0.6331
0.8338
0.9452
0.9877
0.9983
0.9999
0.0010
0.0107
0.0547
0.1719
0.3770
0.6230
0.8281
0.9453
0.9893
0.9990
0.0001
0.0017
0.0123
0.0548
0.1662
0.3669
0.6177
0.8327
0.9536
0.9940
0.0000
0.0001
0.0016
0.0106
0.0473
0.1503
0.3504
0.6172
0.8507
0.9718
0.0000
0.0000
0.0004
0.0035
0.0197
0.0781
0.2241
0.4744
0.7560
0.9437
0.0000
0.0000
0.0001
0.0009
0.0064
0.0328
0.1209
0.3222
0.6242
0.8926
0.0000
0.0000
0.0000
0.0000
0.0001
0.0016
0.0128
0.0702
0.2639
0.6513
0.0000
0.0000
0.0000
0.0000
0.0000
0.0001
0.0010
0.0115
0.0861
0.4013
Page 16 of 19
The Binomial Distribution
© David W. Sabo (1999)
Cumulative Binomial Probabilities (continued)
p=
n= x=
0.05
0.1
0.2
0.25
0.3
0.4
0.5
0.6
0.7
0.75
0.8
0.9
0.95
11
0
1
2
3
4
5
6
7
8
9
10
0.5688
0.8981
0.9848
0.9984
0.9999
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
0.3138
0.6974
0.9104
0.9815
0.9972
0.9997
1.0000
1.0000
1.0000
1.0000
1.0000
0.0859
0.3221
0.6174
0.8389
0.9496
0.9883
0.9980
0.9998
1.0000
1.0000
1.0000
0.0422
0.1971
0.4552
0.7133
0.8854
0.9657
0.9924
0.9988
0.9999
1.0000
1.0000
0.0198
0.1130
0.3127
0.5696
0.7897
0.9218
0.9784
0.9957
0.9994
1.0000
1.0000
0.0036
0.0302
0.1189
0.2963
0.5328
0.7535
0.9006
0.9707
0.9941
0.9993
1.0000
0.0005
0.0059
0.0327
0.1133
0.2744
0.5000
0.7256
0.8867
0.9673
0.9941
0.9995
0.0000
0.0007
0.0059
0.0293
0.0994
0.2465
0.4672
0.7037
0.8811
0.9698
0.9964
0.0000
0.0000
0.0006
0.0043
0.0216
0.0782
0.2103
0.4304
0.6873
0.8870
0.9802
0.0000
0.0000
0.0001
0.0012
0.0076
0.0343
0.1146
0.2867
0.5448
0.8029
0.9578
0.0000
0.0000
0.0000
0.0002
0.0020
0.0117
0.0504
0.1611
0.3826
0.6779
0.9141
0.0000
0.0000
0.0000
0.0000
0.0000
0.0003
0.0028
0.0185
0.0896
0.3026
0.6862
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0001
0.0016
0.0152
0.1019
0.4312
12
0
1
2
3
4
5
6
7
8
9
10
11
0.5404
0.8816
0.9804
0.9978
0.9998
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
0.2824
0.6590
0.8891
0.9744
0.9957
0.9995
0.9999
1.0000
1.0000
1.0000
1.0000
1.0000
0.0687
0.2749
0.5583
0.7946
0.9274
0.9806
0.9961
0.9994
0.9999
1.0000
1.0000
1.0000
0.0317
0.1584
0.3907
0.6488
0.8424
0.9456
0.9857
0.9972
0.9996
1.0000
1.0000
1.0000
0.0138
0.0850
0.2528
0.4925
0.7237
0.8822
0.9614
0.9905
0.9983
0.9998
1.0000
1.0000
0.0022
0.0196
0.0834
0.2253
0.4382
0.6652
0.8418
0.9427
0.9847
0.9972
0.9997
1.0000
0.0002
0.0032
0.0193
0.0730
0.1938
0.3872
0.6128
0.8062
0.9270
0.9807
0.9968
0.9998
0.0000
0.0003
0.0028
0.0153
0.0573
0.1582
0.3348
0.5618
0.7747
0.9166
0.9804
0.9978
0.0000
0.0000
0.0002
0.0017
0.0095
0.0386
0.1178
0.2763
0.5075
0.7472
0.9150
0.9862
0.0000
0.0000
0.0000
0.0004
0.0028
0.0143
0.0544
0.1576
0.3512
0.6093
0.8416
0.9683
0.0000
0.0000
0.0000
0.0001
0.0006
0.0039
0.0194
0.0726
0.2054
0.4417
0.7251
0.9313
0.0000
0.0000
0.0000
0.0000
0.0000
0.0001
0.0005
0.0043
0.0256
0.1109
0.3410
0.7176
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0002
0.0022
0.0196
0.1184
0.4596
13
0
1
2
3
4
5
6
7
8
9
10
11
12
0.5133
0.8646
0.9755
0.9969
0.9997
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
0.2542
0.6213
0.8661
0.9658
0.9935
0.9991
0.9999
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
0.0550
0.2336
0.5017
0.7473
0.9009
0.9700
0.9930
0.9988
0.9998
1.0000
1.0000
1.0000
1.0000
0.0238
0.1267
0.3326
0.5843
0.7940
0.9198
0.9757
0.9944
0.9990
0.9999
1.0000
1.0000
1.0000
0.0097
0.0637
0.2025
0.4206
0.6543
0.8346
0.9376
0.9818
0.9960
0.9993
0.9999
1.0000
1.0000
0.0013
0.0126
0.0579
0.1686
0.3530
0.5744
0.7712
0.9023
0.9679
0.9922
0.9987
0.9999
1.0000
0.0001
0.0017
0.0112
0.0461
0.1334
0.2905
0.5000
0.7095
0.8666
0.9539
0.9888
0.9983
0.9999
0.0000
0.0001
0.0013
0.0078
0.0321
0.0977
0.2288
0.4256
0.6470
0.8314
0.9421
0.9874
0.9987
0.0000
0.0000
0.0001
0.0007
0.0040
0.0182
0.0624
0.1654
0.3457
0.5794
0.7975
0.9363
0.9903
0.0000
0.0000
0.0000
0.0001
0.0010
0.0056
0.0243
0.0802
0.2060
0.4157
0.6674
0.8733
0.9762
0.0000
0.0000
0.0000
0.0000
0.0002
0.0012
0.0070
0.0300
0.0991
0.2527
0.4983
0.7664
0.9450
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0001
0.0009
0.0065
0.0342
0.1339
0.3787
0.7458
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0003
0.0031
0.0245
0.1354
0.4867
14
0
1
2
3
4
5
6
7
8
9
10
11
12
13
0.4877
0.8470
0.9699
0.9958
0.9996
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
0.2288
0.5846
0.8416
0.9559
0.9908
0.9985
0.9998
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
0.0440
0.1979
0.4481
0.6982
0.8702
0.9561
0.9884
0.9976
0.9996
1.0000
1.0000
1.0000
1.0000
1.0000
0.0178
0.1010
0.2811
0.5213
0.7415
0.8883
0.9617
0.9897
0.9978
0.9997
1.0000
1.0000
1.0000
1.0000
0.0068
0.0475
0.1608
0.3552
0.5842
0.7805
0.9067
0.9685
0.9917
0.9983
0.9998
1.0000
1.0000
1.0000
0.0008
0.0081
0.0398
0.1243
0.2793
0.4859
0.6925
0.8499
0.9417
0.9825
0.9961
0.9994
0.9999
1.0000
0.0001
0.0009
0.0065
0.0287
0.0898
0.2120
0.3953
0.6047
0.7880
0.9102
0.9713
0.9935
0.9991
0.9999
0.0000
0.0001
0.0006
0.0039
0.0175
0.0583
0.1501
0.3075
0.5141
0.7207
0.8757
0.9602
0.9919
0.9992
0.0000
0.0000
0.0000
0.0002
0.0017
0.0083
0.0315
0.0933
0.2195
0.4158
0.6448
0.8392
0.9525
0.9932
0.0000
0.0000
0.0000
0.0000
0.0003
0.0022
0.0103
0.0383
0.1117
0.2585
0.4787
0.7189
0.8990
0.9822
0.0000
0.0000
0.0000
0.0000
0.0000
0.0004
0.0024
0.0116
0.0439
0.1298
0.3018
0.5519
0.8021
0.9560
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0002
0.0015
0.0092
0.0441
0.1584
0.4154
0.7712
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0004
0.0042
0.0301
0.1530
0.5123
15
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
0.4633
0.8290
0.9638
0.9945
0.9994
0.9999
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
0.2059
0.5490
0.8159
0.9444
0.9873
0.9978
0.9997
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
0.0352
0.1671
0.3980
0.6482
0.8358
0.9389
0.9819
0.9958
0.9992
0.9999
1.0000
1.0000
1.0000
1.0000
1.0000
0.0134
0.0802
0.2361
0.4613
0.6865
0.8516
0.9434
0.9827
0.9958
0.9992
0.9999
1.0000
1.0000
1.0000
1.0000
0.0047
0.0353
0.1268
0.2969
0.5155
0.7216
0.8689
0.9500
0.9848
0.9963
0.9993
0.9999
1.0000
1.0000
1.0000
0.0005
0.0052
0.0271
0.0905
0.2173
0.4032
0.6098
0.7869
0.9050
0.9662
0.9907
0.9981
0.9997
1.0000
1.0000
0.0000
0.0005
0.0037
0.0176
0.0592
0.1509
0.3036
0.5000
0.6964
0.8491
0.9408
0.9824
0.9963
0.9995
1.0000
0.0000
0.0000
0.0003
0.0019
0.0093
0.0338
0.0950
0.2131
0.3902
0.5968
0.7827
0.9095
0.9729
0.9948
0.9995
0.0000
0.0000
0.0000
0.0001
0.0007
0.0037
0.0152
0.0500
0.1311
0.2784
0.4845
0.7031
0.8732
0.9647
0.9953
0.0000
0.0000
0.0000
0.0000
0.0001
0.0008
0.0042
0.0173
0.0566
0.1484
0.3135
0.5387
0.7639
0.9198
0.9866
0.0000
0.0000
0.0000
0.0000
0.0000
0.0001
0.0008
0.0042
0.0181
0.0611
0.1642
0.3518
0.6020
0.8329
0.9648
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0003
0.0022
0.0127
0.0556
0.1841
0.4510
0.7941
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0001
0.0006
0.0055
0.0362
0.1710
0.5367
© David W. Sabo (1999)
The Binomial Distribution
Page 17 of 19
Cumulative Binomial Probabilities (continued)
p=
n= x=
0.05
0.1
0.2
0.25
0.3
0.4
0.5
0.6
0.7
0.75
0.8
0.9
0.95
16
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
0.4401
0.8108
0.9571
0.9930
0.9991
0.9999
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
0.1853
0.5147
0.7892
0.9316
0.9830
0.9967
0.9995
0.9999
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
0.0281
0.1407
0.3518
0.5981
0.7982
0.9183
0.9733
0.9930
0.9985
0.9998
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
0.0100
0.0635
0.1971
0.4050
0.6302
0.8103
0.9204
0.9729
0.9925
0.9984
0.9997
1.0000
1.0000
1.0000
1.0000
1.0000
0.0033
0.0261
0.0994
0.2459
0.4499
0.6598
0.8247
0.9256
0.9743
0.9929
0.9984
0.9997
1.0000
1.0000
1.0000
1.0000
0.0003
0.0033
0.0183
0.0651
0.1666
0.3288
0.5272
0.7161
0.8577
0.9417
0.9809
0.9951
0.9991
0.9999
1.0000
1.0000
0.0000
0.0003
0.0021
0.0106
0.0384
0.1051
0.2272
0.4018
0.5982
0.7728
0.8949
0.9616
0.9894
0.9979
0.9997
1.0000
0.0000
0.0000
0.0001
0.0009
0.0049
0.0191
0.0583
0.1423
0.2839
0.4728
0.6712
0.8334
0.9349
0.9817
0.9967
0.9997
0.0000
0.0000
0.0000
0.0000
0.0003
0.0016
0.0071
0.0257
0.0744
0.1753
0.3402
0.5501
0.7541
0.9006
0.9739
0.9967
0.0000
0.0000
0.0000
0.0000
0.0000
0.0003
0.0016
0.0075
0.0271
0.0796
0.1897
0.3698
0.5950
0.8029
0.9365
0.9900
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0002
0.0015
0.0070
0.0267
0.0817
0.2018
0.4019
0.6482
0.8593
0.9719
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0001
0.0005
0.0033
0.0170
0.0684
0.2108
0.4853
0.8147
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0001
0.0009
0.0070
0.0429
0.1892
0.5599
20
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
0.3585
0.7358
0.9245
0.9841
0.9974
0.9997
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
0.1216
0.3917
0.6769
0.8670
0.9568
0.9887
0.9976
0.9996
0.9999
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
0.0115
0.0692
0.2061
0.4114
0.6296
0.8042
0.9133
0.9679
0.9900
0.9974
0.9994
0.9999
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
0.0032
0.0243
0.0913
0.2252
0.4148
0.6172
0.7858
0.8982
0.9591
0.9861
0.9961
0.9991
0.9998
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
0.0008
0.0076
0.0355
0.1071
0.2375
0.4164
0.6080
0.7723
0.8867
0.9520
0.9829
0.9949
0.9987
0.9997
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
0.0000
0.0005
0.0036
0.0160
0.0510
0.1256
0.2500
0.4159
0.5956
0.7553
0.8725
0.9435
0.9790
0.9935
0.9984
0.9997
1.0000
1.0000
1.0000
1.0000
0.0000
0.0000
0.0002
0.0013
0.0059
0.0207
0.0577
0.1316
0.2517
0.4119
0.5881
0.7483
0.8684
0.9423
0.9793
0.9941
0.9987
0.9998
1.0000
1.0000
0.0000
0.0000
0.0000
0.0000
0.0003
0.0016
0.0065
0.0210
0.0565
0.1275
0.2447
0.4044
0.5841
0.7500
0.8744
0.9490
0.9840
0.9964
0.9995
1.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0003
0.0013
0.0051
0.0171
0.0480
0.1133
0.2277
0.3920
0.5836
0.7625
0.8929
0.9645
0.9924
0.9992
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0002
0.0009
0.0039
0.0139
0.0409
0.1018
0.2142
0.3828
0.5852
0.7748
0.9087
0.9757
0.9968
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0001
0.0006
0.0026
0.0100
0.0321
0.0867
0.1958
0.3704
0.5886
0.7939
0.9308
0.9885
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0001
0.0004
0.0024
0.0113
0.0432
0.1330
0.3231
0.6083
0.8784
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0003
0.0026
0.0159
0.0755
0.2642
0.6415
25
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
0.2774
0.6424
0.8729
0.9659
0.9928
0.9988
0.9998
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
0.0718
0.2712
0.5371
0.7636
0.9020
0.9666
0.9905
0.9977
0.9995
0.9999
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
0.0038
0.0274
0.0982
0.2340
0.4207
0.6167
0.7800
0.8909
0.9532
0.9827
0.9944
0.9985
0.9996
0.9999
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
0.0008
0.0070
0.0321
0.0962
0.2137
0.3783
0.5611
0.7265
0.8506
0.9287
0.9703
0.9893
0.9966
0.9991
0.9998
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
0.0001
0.0016
0.0090
0.0332
0.0905
0.1935
0.3407
0.5118
0.6769
0.8106
0.9022
0.9558
0.9825
0.9940
0.9982
0.9995
0.9999
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
0.0000
0.0001
0.0004
0.0024
0.0095
0.0294
0.0736
0.1536
0.2735
0.4246
0.5858
0.7323
0.8462
0.9222
0.9656
0.9868
0.9957
0.9988
0.9997
0.9999
1.0000
1.0000
1.0000
1.0000
1.0000
0.0000
0.0000
0.0000
0.0001
0.0005
0.0020
0.0073
0.0216
0.0539
0.1148
0.2122
0.3450
0.5000
0.6550
0.7878
0.8852
0.9461
0.9784
0.9927
0.9980
0.9995
0.9999
1.0000
1.0000
1.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0001
0.0003
0.0012
0.0043
0.0132
0.0344
0.0778
0.1538
0.2677
0.4142
0.5754
0.7265
0.8464
0.9264
0.9706
0.9905
0.9976
0.9996
0.9999
1.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0001
0.0005
0.0018
0.0060
0.0175
0.0442
0.0978
0.1894
0.3231
0.4882
0.6593
0.8065
0.9095
0.9668
0.9910
0.9984
0.9999
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0002
0.0009
0.0034
0.0107
0.0297
0.0713
0.1494
0.2735
0.4389
0.6217
0.7863
0.9038
0.9679
0.9930
0.9992
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0001
0.0004
0.0015
0.0056
0.0173
0.0468
0.1091
0.2200
0.3833
0.5793
0.7660
0.9018
0.9726
0.9962
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0001
0.0005
0.0023
0.0095
0.0334
0.0980
0.2364
0.4629
0.7288
0.9282
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0002
0.0012
0.0072
0.0341
0.1271
0.3576
0.7226
Page 18 of 19
The Binomial Distribution
© David W. Sabo (1999)
Cumulative Binomial Probabilities (continued)
p=
n= x=
30
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
0.05
0.1
0.2
0.25
0.2146
0.5535
0.8122
0.9392
0.9844
0.9967
0.9994
0.9999
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
0.0424
0.1837
0.4114
0.6474
0.8245
0.9268
0.9742
0.9922
0.9980
0.9995
0.9999
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
0.0012
0.0105
0.0442
0.1227
0.2552
0.4275
0.6070
0.7608
0.8713
0.9389
0.9744
0.9905
0.9969
0.9991
0.9998
0.9999
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
0.0002
0.0020
0.0106
0.0374
0.0979
0.2026
0.3481
0.5143
0.6736
0.8034
0.8943
0.9493
0.9784
0.9918
0.9973
0.9992
0.9998
0.9999
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
© David W. Sabo (1999)
0.3
0.0000
0.0003
0.0021
0.0093
0.0302
0.0766
0.1595
0.2814
0.4315
0.5888
0.7304
0.8407
0.9155
0.9599
0.9831
0.9936
0.9979
0.9994
0.9998
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
0.4
0.0000
0.0000
0.0000
0.0003
0.0015
0.0057
0.0172
0.0435
0.0940
0.1763
0.2915
0.4311
0.5785
0.7145
0.8246
0.9029
0.9519
0.9788
0.9917
0.9971
0.9991
0.9998
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
0.5
0.0000
0.0000
0.0000
0.0000
0.0000
0.0002
0.0007
0.0026
0.0081
0.0214
0.0494
0.1002
0.1808
0.2923
0.4278
0.5722
0.7077
0.8192
0.8998
0.9506
0.9786
0.9919
0.9974
0.9993
0.9998
1.0000
1.0000
1.0000
1.0000
1.0000
0.6
0.7
0.75
0.8
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0002
0.0009
0.0029
0.0083
0.0212
0.0481
0.0971
0.1754
0.2855
0.4215
0.5689
0.7085
0.8237
0.9060
0.9565
0.9828
0.9943
0.9985
0.9997
1.0000
1.0000
1.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0002
0.0006
0.0021
0.0064
0.0169
0.0401
0.0845
0.1593
0.2696
0.4112
0.5685
0.7186
0.8405
0.9234
0.9698
0.9907
0.9979
0.9997
1.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0001
0.0002
0.0008
0.0027
0.0082
0.0216
0.0507
0.1057
0.1966
0.3264
0.4857
0.6519
0.7974
0.9021
0.9626
0.9894
0.9980
0.9998
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0001
0.0002
0.0009
0.0031
0.0095
0.0256
0.0611
0.1287
0.2392
0.3930
0.5725
0.7448
0.8773
0.9558
0.9895
0.9988
The Binomial Distribution
0.9
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0001
0.0005
0.0020
0.0078
0.0258
0.0732
0.1755
0.3526
0.5886
0.8163
0.9576
0.95
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0001
0.0006
0.0033
0.0156
0.0608
0.1878
0.4465
0.7854
Page 19 of 19