MATH 2441
Probability and Statistics for Biological Sciences
Calculating Probabilities: I
Equally-likely Simple Events and Branching Diagrams
Some of the most complicated problems in all of mathematics has to do with the determination of
probabilities. However, most of our work with probabilities in this course falls into just a few well-defined
situations which can be handled using one of a few relatively simple approaches.
Recall the long-run relative frequency definition of probability described in the preceding document:
Pr( A) 
m
N
where
N = the number of times the experiment is repeated
and
m = the number of times the event A occurs in those N repetitions
This defines Pr(A) well if as N becomes very large in principle, the ratio m/N becomes more and more
constant (apologies for the non-technical language here!).
This definition can be exploited in a variety of ways to come up with probabilities of practical value. In this
document, we look at a few simple examples to introduce some useful methods.
Equally-Likely Simple Events
In many situations, the outcomes of an experiment can be expressed in terms of a set of simple events
which are all equally likely because of the nature of the experiment. In such a case, if there are N equally
likely simple events, then we know that each one of them must have a probability of 1/N. Furthermore, if we
have a compound event A which corresponds to m distinct simple events, then Pr(A) = m/N (the sum of m
terms, each equal to 1/N). This reduces the problem to one of counting how many simple events there are
in A -- not necessarily an easy task in some cases -- but at least something to start with.
The counting process is often facilitated by what we will call a branching diagram, which illustrates the tally
of events in an experiment. We will illustrate the basic methodology with several simple examples.
Example 1: Tossing Fair Coins
By a "fair coin" we mean one which has the same
likelihood of falling tails as of falling heads. To develop
a list of all possible outcomes of an experiment in which
two distinct coins are tossed, we proceed as follows.
Outcomes
H
HH
T
HT
H
TH
T
TT
H
The first coin can fall either heads up (H) or tails up (T)
 the two outcomes having the same likelihood. When
the second coin is tossed, it can likewise result in H or
T, with equal likelihood. This two-coin experiment is
pictured in the branching diagram to the right. Writing
the overall outcome of the experiment as a string of H's
and T's to indicate how each coin in the sequence fell,
we see from this picture that there is exactly four distinct
outcomes to an experiment in which two coins are
flipped:
T
First Coin
Second Coin
HH meaning the first coin falls heads up and the second coin falls heads up
HT meaning the first coin falls heads up and the second coin falls tails up
TH meaning the first coin falls tails up and the second coin falls heads up
David W. Sabo (1999)
Calculating Probabilities: I
Page 1 of 11
and
TT meaning the first coin falls tails up and the second coin falls tails up.
How do we know there are just these four outcomes and that we've gotten all possible outcomes? Well, the
first coin can fall only H or T, no other possibilities for it. In this diagram, this shows up as the list with the
two possibilities {H, T} in a vertical column above the label "first coin." The dotted branching lines on the
extreme left of the diagram indicate that these two outcomes are the only ones that can happen in passing
from the state "no coins flipped" to the state "the first coin is flipped."
Then, whatever the outcome for the first coin, the second can either fall H or T. In the diagram, this is
shown by attaching an {H, T} branch to each of the two possible outcomes for the first coin.
Each path through the diagram from "no coin flipped" to an outcome for the last coin flipped corresponds to
a way in which the overall two-coin-flip experiment can occur. In fact, each outcome in the right-most
column corresponds to a distinct way in which the two-coin experiment can occur. Because of the way this
diagram is constructed, we know that there are no other ways the two-coin experiment can occur.
Furthermore, at each branch point, the probability of the two paths are equal (because of our definition of a
"fair" coin). Thus, each path through this diagram has the same probability. Thus, the four simple events
{HH, HT, TH, TT} are equally likely, and so each must have a probability of 1/4 or 0.25:
Pr(HH) = 0.25,
Pr(HT) = 0.25,
Pr(TH) = 0.25,
Pr(TT) = 0.25
It is now easy to see how to obtain probabilities for
an experiment in which we flip three coins. We
just need to attach an {H, T} branch to each
outcome for the second coin above, which gives
the diagram to the right. Every outcome of the
two-coin experiment leads to two possible
outcomes in a three-coin experiment, for a total of
eight. Since each path at a branch is equallylikely for fair coins, the eight distinct outcomes of
the three-coin experiment are themselves equally
likely, and so each has a probability of 1/8. Thus,
we can write
HHH
H
H
T
HHT
H
HTH
T
HTT
H
THH
T
THT
H
TTH
T
TTT
T
H
T
Pr(HHH) = 1/8 or 0.125
Pr(HHT) = 1/8 or 0.125,
T
etc.
H
First Coin
Second Coin
Third Coin
Each of the eight outcomes, {HHH, HHT, HTH,
HTT, THH, THT, TTH, TTT} are simple events.
We cannot express any of them as a set of one or more of the other seven. They are also mutually
exclusive, since if the three coins fall in one of the eight patterns, they cannot have fallen in any of the other
seven patterns.
An example of a compound event is the event:
A = two coins fall heads up when three coins are flipped
because
A = {HHT, HTH, THH}
That is, A represents a set of three of the simpler events from our list. From the third property we listed for
probabilities in the previous document, we know that since HHT, HTH, and THH are mutually exclusive
simple events, each with probability 0.125, then
Pr(A) = Pr(HHT) + Pr(HTH) + Pr(THH) = 0.125 + 0.125 + 0.125 = 0.375
The probability of getting two coins heads up when three coins are flipped is 0.375. This means that if we
flip a set of three coins repeated, then approximately 37.5% of the time, the coins will fall showing two
heads.
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Calculating Probabilities: I
David W. Sabo (1999)
In this same way, the probability can now be calculated for any event B which can be expressed as some
collection of the eight simple events listed above.
We could continue the process to experiments in which four coins are flipped (for which there would be 2 4 =
16 possible outcomes), and so on. All that changes is that the diagram, and hence the list of outcomes
becomes longer and longer. Later in the course, we will present some formulas or procedures that simplify
the process. This coin toss experiment is a special case of such an important type of random experiment
that it has its own name: the binomial experiment. It has important applications in many areas of
biological sciences.

Example 2: The Selection of Contestants (or… the unfair coin!)
Let us suppose that three individuals, 15-years older or older, are to be selected at random from the
population of Burnaby to win some prize. When the winners are announced, it is found that they are Larry,
aged 24; Curley, aged 55, and Moe, aged 63 -- that is, two of the three successful contestants are 55 years
old or older. Accusations that the contest was rigged in favor of senior citizens are made. Assuming that no
one claims to have seen the organizers overtly tampering with the selection process, is there any way to
decide whether further investigation is necessary here?
S
SSS
Solution
S
One way to decide if there is any value in investigating the
Y
SSY
S
charge of rigging the selection further is to determine how
S
SYS
unlikely it might be that two of three contestants would be 55
Y
years old or older if everyone in Burnaby had an equal
Y
SYY
likelihood of being selected (that is, the selection process was
S
YSS
fair). We can view the contestant selection process as a little
S
like the process of flipping a coin as far as age categories are
Y
YSY
concerned. We will let S stand for the event that a contestant
Y
S
YYS
is 55 years old or older, and let Y stand for the event that the
Y
First
selected contestant is younger than 55 years. Then the first
Contestant
Y
YYY
contestant selected will either be S or Y, and whatever the first
Second
Third
contestant is, the second can also either be S or Y, and
Contestant
Contestant
similarly the third will also be either S or Y. In fact, you can see
that the possible ways of choosing three contestants (as far as their S-ness or Y-ness is concerned) is given
by the same kind of branching diagram as used about for the three-coin experiment, except that the letters H
and T are replaced by S and Y, as shown to the right.
Of the eight ways in which the contestants could have turned out (as far as their S or Y characteristics are
concerned), three of them involve the selection of two S's and one Y -- exactly what happened in this
contest. We need to figure out probabilities for each of these three outcomes.
In the case of flipping a fair coin, we knew that H and T occurred with equal probability for each coin, and so
we were able to conclude that all eight final outcomes in the branching diagram were equally-likely, and
hence each had a probability of 1/8 or 0.125. In this example, there is no reason to assume that in selecting
a person at random from Burnaby's 15 or over population, we will be equally likely to get an S or a Y. In
fact, consulting Government of British Columbia statistical data for 1997 (the latest year available at the time
of writing this), we find estimates that of the 157,742 persons 15 years or older in Burnaby that year, there
are 116,606 persons (or 73.9%) between 15 and 54 years old and 41,136 persons (or 26.1%) 55 years or
older. Since probabilities are the same thing as relative frequencies, we can equate these percentages for
each group with the probability that a person of that group will be selected in a truly fair random draw.
In fact, if we repeated selecting trios of contestants at random from the population of Burnaby, we would
have found that 73.9% of the first contestants selected would be Y, and 26.1% of the first contests would be
S. Of the 73.9% of the trios in which the first contestant is Y, 73.9% would have the second contestant a Y.
Thus the proportion of trios in which the first two contestants were Y's would be 73.9% of 73.9% or (0.739) 2.
Continuing in this way, we obtain the following percentages for each of the eight combinations of contestants
that can result:
David W. Sabo (1999)
Calculating Probabilities: I
Page 3 of 11
26.1%
S
26.1%
73.9%
S
26.1%
73.9%
26.1%
Y 73.9%
73.9%
26.1%
S
26.1%
73.9%
Y
73.9%
Y
26.1%
73.9%
First
Contestant
Second
Contestant
S
SSS
(0.261)(0.261)(0.261)
Y
SSY
(0.261)(0.261)(0.739)
S
SYS
(0.261)(0.739)(0.261)
Y
S
SYY
YSS
(0.261)(0.739)(0.739)
Y
YSY
(0.739)(0.261)(0.739)
S
YYS
(0.739)(0.739)(0.261)
Y
YYY
(0.739)(0.739)(0.739)
Third
Contestant
(0.739)(0.261)(0.261)
The products to the right indicate the fraction of a large number of trios of randomly selected contestants
that will have the indicated selection history. We can summarize this information in a table:
contestant
selection:
SSS
SSY
SYS
SYY
YSS
YSY
YYS
YYY
proportion of a large number of trios of
contestants with this pattern:
(0.261)(0.261)(0.261)
(0.261)(0.261)(0.739)
(0.261)(0.739)(0.261)
(0.261)(0.739)(0.739)
(0.739)(0.261)(0.261)
(0.739)(0.261)(0.739)
(0.739)(0.739)(0.261)
(0.739)(0.739)(0.739)
=
=
=
=
=
=
=
=
0.0178
0.0503
0.0503
0.1425
0.0503
0.1425
0.1425
0.4036
Now we can ask: what is the probability that a truly random selection of three contestants would have
resulted in two S's and one Y? From the table, we see that in the long run, if many trios of three Burnaby
residents are selected randomly, we would expect that a proportion 0.0503 + 0.0503 + 0.0503 = 0.1509
would be of such a type (just add up the proportions which are SSY, SYS, or YSS). Thus, we can say that
the probability of getting two S's and one Y in a randomly selected group of three individuals is 0.1509, or
very close to 15%.
Is this number small enough to cause us to doubt such an event could happen? Most people would agree
"not really." You will see as the course progresses that the conventional meaning of "rare occurrence" is
usually taken to be one with a probability of less than 5%, and in some cases, even much smaller. If we had
found that the probability of obtaining the combination of two S's and one Y in a truly random selection
process was, say, only 0.001, we would have grounds for doubting that a truly random selection had
occurred. In that case, the fact that the event actually occurred casts doubt in most people that the
probability of occurrence could really be as small as 0.001. However, there are many things happen to each
of us each day that have probabilities of 15% and we are not surprised. So, the conclusion here has to be
that there really isn't evidence on the face of things to support a claim of tampering with the random
selection process.

This last example gives rise to a number of aspects of probability calculations which are worth noting at this
point, even though more detailed treatment of some must be put off until later in the course:
(i.)
The 15% probability obtained in this example seems to give more of a "non-answer" than an
answer to the original question of whether the contestants resulted from a truly fair or random
Page 4 of 11
Calculating Probabilities: I
David W. Sabo (1999)
selection process. The probability isn't so small that we find it difficult to believe the event could
have happened, yet it isn't so large that it seems to indicate the observed outcome is highly likely.
It would still be possible to detect an unfair selection process if there was a succession of such
contests in which two S's and one Y were picked consistently. What this result means is that over
the long run, 15% of such selections should result in two S's and one Y. Thus, for a pair of
contests, 15% will have the first group of contestants made up of two S's and one Y, and 15% of
these will have the second group made up of two S's and one Y. Thus, the probability of a two
contests in a row each involving two S's and one Y is just (0.1509) 2  0.023 or about one in 40.
This is a fairly low probability, and you might be justified in having some suspicion of unfairness if
two contests in a row involved two S's and one Y. Now, if you looked at three successive contests,
then the probability of all three involving two S's and one Y would be (0.1509) 3  0.0034, or about
one in 300. To observe such an outcome for three successive contests is thus quite surprising,
and probably means the selection process should be investigated.
Would the occurrence of three consecutive contests involving two S's and one Y mean that the
organizers are deliberately cheating? Not necessarily. These outcomes could just be the result of
an incredible string of coincidences. Or, it could be the result of an unwittingly faulty selection
process. Our calculations are based on the assumption that all 157,742 residents of Burnaby have
the same likelihood of being selected. But, there may be all sorts of reasons why a convenient
contestant selection process might inadvertently favor those who are 55 years old or older. For
example, perhaps contestants are selected by first selecting a random telephone number, and then
phoning them in the middle of the day. Residents under the age of 55 may be less likely to be
home to receive such a call, and so less likely to be selected as contestants. You can probably
think of other procedures that might favor residents in the S category over residents in the Y
category. This may not be such a big deal in selecting contestants for a contest perhaps, but such
sampling biases can result in seriously incorrect conclusions in scientific and technical research.
Ways to avoid such errors fall under the topics of experimental design and sampling theory.
(ii.)
Make sure that you see that Example 2 is just a slight generalization of Example 1, and that you
can explain why we referred to Example 2 as "the unfair coin" example. By "unfair coin", we mean
a coin which is more likely to fall with one side up than with the other.
Situations such as these two, which consist of a fixed number of repetitions (or trials) of the same
process, and that process can result in one of just two possible simple outcomes, and the
probability of each simple outcome is the same for all trials, is a pattern that recurs in a large
number of scientific and technical situations. The process is so common that it has been given its
own special name: the binomial experiment. The binomial experiment has been studied
extensively, and we will look at it more generally and in more detail later in the course.
(iii.)
The calculations in Example 2 show some obvious patterns. The probabilities associated with each
of the eight simple outcomes are quite easy to set up. We have that for the selection of any one
contestant, Pr(Y) = 0.739 and Pr(S) = 0.261. Then, the probability of any given sequence of S's
and Y's is just the product of 0.739 for each S and 0.261 for each Y. This occurs because every
time we take an S branch in the diagram, the proportion of occurrences is reduced by a factor of
0.261, and every time we take a Y branch in the diagram, the proportion of occurrences is reduced
by a factor of 0.739.
Furthermore, if we ask for the probability of, say, selecting two S's and one Y without regard to their
order of selection, we see that we are really asking for the probability of occurrence of one of the
three simple outcomes {SSY, SYS, YSS}. Notice that these three outcomes amount to all the
possible distinct arrangements of two S's and one Y. It's quite easy to convince yourself that there
are only three distinct arrangements of two S's and one Y -- either use the branching diagram
above, or spend a few minutes trying to think of a fourth distinct arrangement. We will present a
formula later for working out this number in more complicated cases where intuition or drawing a
branching diagram may not be adequate or convenient.
(iv.)
Finally, a comment about using the numbers Pr(Y) = 0.739 and Pr(S) = 0.261. The basic definition
of a probability as a relative frequency justified this. We had the information that there were
157,742 residents of Burnaby in 1997 of which 116,606 were Y's. Thus, the Y's had a relative
frequency of 116,606/157,742  0.739. Assuming that the original population figures are accurate
(and one should have some reservations perhaps -- how could you count over 116,000 people right
down to the last individual, as implied by the 6 in the last figure?), identification of population
David W. Sabo (1999)
Calculating Probabilities: I
Page 5 of 11
proportions with probabilities is valid.
More usually, we don't have exact values for population numbers. We would then have to estimate
population proportions using information about proportions in random samples of that population.
Thus, another source of potential error in conclusions would be possible sampling errors in getting
estimates of these population proportions.
Finally, there is one somewhat more subtle issue in connection with these probabilities that needs
mentioning, although it is of no importance at all in this particular example. We assumed that the
correct composition of the population was 116,606 Y's and 41,136 S's for a total of 157,742 people
in all. If a person is picked at random from this population, then the probability of picking a Y is
given by 116,606/157,742. We used this probability for the second and third contestants selected
as well. However, if the three contestants must be distinct individuals, then the numbers in the
populations from which the second and third contestants are selected are not the same as for the
population from which the first contestant is selected, and so the probabilities are not really the
same either. For instance, if the first contestant selected is a Y, then the second contestant will be
selected from a population of 116,605 Y's out of a total of 157,741 people -- the first contestant has
been removed from the list of people eligible to be the second contestant. So, for the second
contestant, Pr(Y) is really 116,605/157,741 rather than 116,606/157,742. In this case, the
difference between these two proportions is certainly negligible, but that would not be true if we
were drawing items from a very small population. We speak of this being sampling without
replacement, to indicate that when a person is picked as one contestant, they are not "returned" to
the pool of eligible contestants for the selection of the next contestant. (For the alternative method,
sampling with replacement, an element of the population would be selected, its identity recorded,
and then it would be returned to the population and be eligible for selection again.) The method we
used in Example 2 is exactly correct for a sampling with replacement, and is approximately correct
for sampling without replacement when the sampling process itself doesn't significantly change the
proportions of the population.
After that lengthy discussion of Example 2, we must move along quickly to keep the length of this document
under control. However, we have two more examples that illustrate extensions of, and perhaps alternatives
to, the simple branching diagram in the calculation of probabilities for some basic, but important,
experimental patterns: rolling dice and lotteries.
Example 3: Rolling (Fair) Dice
You are familiar with the usually cube-shaped dice (we're going use the word "dice" to
refer to one or more than one of these things -- some people call a single one a "die"
and more than one "dice", but then surely we get into all sorts of trouble with words like
"rice" and "spice" and "mice", not to mention "ice" ). By a "fair" dice, we mean one for which every face
has the same likelihood of ending up on top when the dice is tossed.
When one dice is tossed, the possible outcomes are just the values {1, 2, 3, 4, 5, 6}. If the dice is fair, then
the six outcomes are equally likely, and so we can write:
Pr(1) = 1/6
Pr(4) = 1/6
Pr(2) = 1/6
Pr(5) = 1/6
Pr(3) = 1/6
Pr(6) = 1/6
So, what is the probability that you will get an even number when you roll one dice? Simply
Pr(even number) = Pr(2 or 4 or 6) = Pr(2) + Pr(4) + Pr(6) = 1/6 +1/6 + 1/6 = 0.5,
since the events {1, 2, 3, 4, 5, 6} are mutually exclusive simple events.
Things get a bit more complicated when we deal with two dice. One way to sort things out is by using a
branching diagram with two columns: the first indicating how the first dice lands, and the second indicating
how the second dice lands. This is shown in abbreviated form below.
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Calculating Probabilities: I
David W. Sabo (1999)
The numbers down the left edge of the figure is the outcome for
the first dice. Only the first two of six possible outcomes is
shown. For each of the six ways the first dice can fall, the
second can fall in one of six ways, as shown in the second
column. Thus, the complete branching diagram would consist of
six sections of six branches each. From this we conclude that
there are 36 distinct ways that two dice can fall. Further, for fair
dice, each outcome has the same likelihood, so that the 36
simple outcomes for two dice are all equally likely. We can also
represent the outcomes for the two-dice experiment as a pair of
numbers in brackets, indicating how the first and second dice fall
respectively.
1 (1, 1)
2 (1, 2)
3 (1, 3)
4 (1, 4)
1
5 (1, 5)
6 (1, 6)
For a three dice experiment, each of the 36 branches in the
figure to the right would split into six more branches, giving a
total of 36 x 6 = 216 possible outcomes.
2
1
2
(2, 1)
3
4
(2, 3)
5
6
Even for a two-dice experiment, the branching diagram is a bit
cumbersome to work with. Thus, it is easy to see even from this
incomplete diagram that
(2, 2)
(2, 4)
(2, 5)
(2, 6)
second dice
6
Pr( two ones) = 1/36
because the outcome (1, 1) can only occur in the first set of branches emerging from the "first dice falls 1"
event on the left. The pattern of two-dice events listed down the right is clear enough that we know for sure
there will not be a (1, 1) lower down as well.
It's somewhat more difficult to determine Pr(sum of dots on both dice = 7). We see two simple events of this
type in the part of the branching diagram that's shown: (1, 6) and (2, 5). With a bit of thought, you should be
able to see that the complete diagram would have four more outcomes corresponding to a sum of 7, namely,
(3, 4), (4, 3), (5, 2) and (6, 1). Thus, Pr(sum = 7) corresponds to any one of six equally likely simple events
and so
Pr(sum = 7) = 6 x (1/36) = 1/6.
To compute probabilities of various sums for the two-dice experiment, it might be simpler to display the
possible outcomes in a more tabular form:
first
dice
1
2
3
4
5
6
1
2
2
3
4
5
6
7
3
4
5
6
7
8
second dice
3
4
4
5
6
7
8
9
5
6
7
8
9
10
5
6
6
7
8
9
10
11
7
8
9
10
11
12
The numbers in the body of the table give the sum of the two dice corresponding to the outcomes for each in
the left-hand column and top row. Each of the 36 entries in the body of the table represent equally-likely
outcomes, so each have a probability of 1/36. Thus, to work out Pr(sum of two dice = 7), we just count up
the number of 7's that show up in the body of this table. Since there are six of them, we reconfirm the result
obtained just above that Pr(sum = 7) = 6/36 or 1/6.
Note that we can describe the possible outcomes of the two-dice experiment in terms of the sum of the two
dice. Described in this way, the list of possible outcomes is {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}. However, only
two of these eleven outcomes are simple events (the others correspond to two or more different ways that
the dice can fall), and they are not equally-likely. The "equally-likeliless" of a set of events must be the
consequence of some known aspect of the experiment.

David W. Sabo (1999)
Calculating Probabilities: I
Page 7 of 11
Example 4: Lotto 649
Very briefly, we look at one more example that is amenable to analysis based on branching diagrams -lotteries based on selecting several of a larger set of numbers. The most familiar version in this area of the
world is probably the Lotto 649. Ticket holders must select six numbers from 1 through 49. The grand prize
is won if those same six numbers are drawn by the lottery operators. Great effort is made to ensure that all
possible six-number combinations have the same likelihood. To work out the probability of winning the
grand prize, we just need to determine how many different six-number combinations can be made up from
the numbers between 1 and 49 inclusive.
We can only show a very small part of the branching diagram in this case, but hopefully it's enough to help
you see the pattern:
2
2
3
5
6
3
4
fifth and sixth
numbers
1
49
first
number:
49 choices
49
1
third
number:
47 choices
left
fourth
number:
46 choices
left
3
4
2
49
49
second number:
48 choices left
Think of the experiment as one in which we select the first number (which can be any one of the 49 possible
numbers), and then the second number (but now, the second number has only 48 possibilities, since one of
the 49 has already been taken in the first selection). Once the first two numbers are selected, there are only
47 candidates left to be the third number. Once the first three numbers are selected, there are only 46
candidates left to be the fourth number. The fifth is chosen from among 45 remaining possibilities, and the
sixth number from among 44 possibilities.
Notice in the set of branches representing the second number, we omit the number that has already been
picked as the first number. In the set of branches representing the selection of the third number, we omit the
two numbers that have already been selected in each case.
Anyway, it is quite easy to see how many branches the entire diagram would have if drawn up completely.
There are 49 sections corresponding to each possible choice of the first number. For each of these 49
choices of a first number, there are 48 choices for the second number. Thus, there are 49 x 48 = 2352 ways
of selecting the first two numbers. For each of these, there are 47 ways of selecting the third number, so the
number of ways of selecting the first three numbers is (49 x 48) x 47 = 110,544. For each of these, there are
46 choices of the fourth number, giving (49 x 48 x 47) x 46 = 5,085,024 ways of choosing the first four
numbers. For each of these, there are 45 ways of choosing the fifth number, giving a total of (49 x 48 x 47 x
46) x 45 = 228,826,080 ways of choosing the first five numbers. Finally, for each of these ways of choosing
the first five numbers, there are 44 ways of choosing the sixth number. Thus, it looks like the number of
different ways in which the six numbers in the Lotto 649 can be chosen is
49 x 48 x 47 x 46 x 45 x 44 = 10,068,347,520
(CP-1)
Fortunately, this is not quite correct or there would be very few winners (though each would win an
extremely big prize). But it is a start.
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Calculating Probabilities: I
David W. Sabo (1999)
The error in the calculation leading to (CP-1) is not too hard to track down. Look at the branching diagram
again. If we followed the top path all the way through the six stages, we'd come up with the selection of the
six numbers {1, 2, 3, 4, 5, 6} in that order. Now, take the top path through the diagram starting with a 2 as
the first number chosen. Then the six numbers chosen would be in order {2, 1, 3, 4, 5, 6}. But this is exactly
the same lottery ticket as the first one. Only the identity of the six numbers chosen is important -- the order
in which you choose the six numbers is irrelevant. So, it looks like our branching diagram is counting some
tickets more than once, and so the number obtained in (CP-1) is too large.
If you think about it for a minute, you'll realize that the ticket with the six numbers {1, 2, 3, 4, 5, 6} will show
up in our list once for every possible order in which these six numbers can be written. In fact, every
combination of six numbers will show up in our list that many times. In how many different orders can you
write six numbers? Once again, imagine a branching diagram. The first number in the sequence can be
any one of the six numbers. Then, for each choice of the first number, the second number in the sequence
can be any of the five remaining numbers. So, there are 6 x 5 = 30 ways to place the first two numbers in
the sequence. The third number can be any of the remaining four numbers after the first two are placed, so
there are (6 x 5) x 4 = 120 distinct sequences for the first three numbers picked. Then the fourth number
can be any of the three numbers not placed so far, meaning that there are (6 x 5 x 4) x 3 = 360 distinct
sequences of the first four numbers. The fifth number placed can be any of the remaining two numbers, so
that there are (6 x 5 x 4 x 3) x 2 = 720 distinct sequences for the first five numbers. Finally, once the first
five numbers have been placed, there is only one choice left for the sixth. Thus, there are (6 x 5 x 4 x 3 x 2)
x 1 = 720 different arrangements of any combination of six numbers. This means that the result (CP-1)
counts every valid Lotto 649 ticket 720 times!.
So, dividing (CP-1) by 720 gives
number of distinct Lotto 649 tickets = 10,068,347,520/720 = 13,983,816
(CP-2)
or just under 14 million. Each of these tickets are equally likely to be prize winners. Thus, the probability of
winning the Lotto 649 is 1/13,983,816 (or about 1 in 14 million). This means that in the long run, there is
about one winning ticket for every 14 million or so tickets sold.
Humans have a hard time understanding either very large numbers (such as 14 million) or very small
numbers (such as 1/14 million). One way to get a sense of what a probability like this might mean in
practice is to look at other events in life. For example, approximately 300 people in the United States are
killed each year by lightning. Assuming the 300 million or so people in the US are all equally likely to be
struck by lighting (which is not entirely accurate because lightning strikes are more frequent in certain parts
of the country than in others), then each person in the US has a probability of about 1 in a million of being
fatally struck by lightning each year -- a roughly fourteen times greater chance than winning the grand prize
with a single Lotto 649 ticket.

Later in the course, we will return to several important features that appear in this Lotto 649 example. We
just note them here briefly.
Factorials, Permutations, Combinations
First, you'll notice that products of successive whole numbers (or integers) occur often in counting up
branches of a diagram such as the one just above. It is convenient to introduce the so-called factorial
function for an integer: it is the product of that integer and all smaller integers down to 1. The exclamation
point is used to indicate a factorial. So, for example,
6! = 6 x 5 x 4 x 3 x 2 x 1 = 720.
We would speak the left-hand side of this equation as "six-factorial". This number, 6!, occurred above when
we calculated the number of ways in which 6 items could be arranged in a row. Such alternative
arrangements of items are called permutations. Thus, one generalization from the work in the last example
is that
the number of permutations of n objects or items = n!
David W. Sabo (1999)
Calculating Probabilities: I
Page 9 of 11
Secondly, recall that in the result (CP-1) above, we got a product of a sequence of consecutive whole
numbers, but the sequence didn't go all the way down to 1, so it wasn't really a factorial. However, such
sequential products of whole numbers can always be written as a quotient of two factorials. In the case of
(CP-1), the product was
49 x 48 x 47 x 46 x 45 x 44
But note the following:
49! = 49 x 48 x 47 x 46 x 45 x 44 x 43 x 42 x ... x 3 x 2 x 1
43!
Thus, we can write
49  48  47  46  45  44 
49!
43!
In fact, this leads to a much more useful formula than just being able to express sequential products of
whole numbers as a quotient of factorials. In the calculation displayed in (CP-1) is really a calculation of
how many ways we can choose 6 things from 49 where the order in which we choose them is important. As
mentioned earlier, in the jargon of probability theory, those ordered groups of things are called
permutations. From that example, we conclude that the number of permutations of six items selected from
a pool of 49 items is given by
P49,6 
49!
49!

43! (49  6)!
where we have rewritten the denominator 43! in terms of the numbers 49 and 6 occurring in the problem.
Thus, it would appear in general that the number of permutations of m items selected from a pool of N items,
denoted by the symbol PN,m, is given by
PN ,m 
N!
( N  m)!
(CP-3)
This formula is quite important in some applications, and we will return to it again before the end of the
course.
Finally, if order of selection is not important, then the above example indicates that we must divide P N,m by
m! to get the number, CN,m, of unique combinations of m things chosen from a pool of N things. That is
C N ,m 
PN , m
m!

N!
( N  m)! m!
(CP-4)
Question
A laboratory has a supply of 20 white rats. Each of five rats selected at random from this pool of 20 is to be
assigned to a specific researcher for observation. In how many distinct ways can this assignment of 5 rats
to 5 researchers be done?
Solution
It is obvious that this question asks how many ways five items (rats) can be chosen from the pool of 20
items. Since each rat will be assigned to a specific researcher, we need to calculate the number of
permutations rather than the number of combinations. Order of assignment is important here, because if
two researchers switch rats, they've really created a new assignment (even when the same five rats overall
are used in the experiment). Thus, the answer to this question is:
Page 10 of 11
Calculating Probabilities: I
David W. Sabo (1999)
P20,5 
20!
20!

 20 19 18 17 16  1,860 ,480
(20  5)! 15!
(Some scientific calculators have function keys for PN,m and CN,m, so that the expansion into factorials and
the expansion of the factorials as done above would not be necessary.)

Question
A laboratory has a supply of 20 white rats. Five of these are to be selected for use in a particular
experiment. In how many different ways can this group of five rats be selected?
Solution:
This is similar to the previous question except that now none of the five selected rats are not going to be
treated in a distinctive way -- so order of selection is not important. Thus, the question asks for how many
combinations of five rats can be chosen from the pool of twenty, which is
C 20,5 
20!
20!

 15,504
(20  5)! 5! 15! 5!
Thus, there are 15,504 distinct samples of 5 rats that can be selected from the pool of 20 available rats.

David W. Sabo (1999)
Calculating Probabilities: I
Page 11 of 11