Binomial Distributions
Requirements for Binomial Distributions
1. There are a fixed number n of observations.
2. The n observations are all independent. That is, knowing the result of one observation does not
change the probabilities we assign to other observations.
3. Each observation falls into one of just two categories, which for convenience we call “success” and
“failure.”
4. The probability of a success, call it p, is the same for each observation.
n = number of observations
p = probability of success
q = 1 – p = probability of a failure
x = number of successes
Binomial probability
P(X = x) = nCx · px · qn-x
The blood types of successive children born to the same parents are independent and have fixed
probabilities that depend on the genetic makeup of the parents. Each child born to a particular set of
parents has probability 0.25 of having blood type O. If these parents have 5 children, what is the probability
that exactly 2 of them have type O blood?
The binompdf command
1. Press 2nd VARS for the DISTR menu. Scroll down to binompdf( and press Enter. This is menu item A if you
have a TI-84 calculator, but it is menu item 0 on a TI-83 calculator.
2. The syntax for the binomial probability density function command is binompdf(n,p,x).
• n: This is the number of trials. For this example, n = 5 (the number of children).
• p: This is the “success” probability. For this example, p = 0.06 (the probability that the child will have
blood type O). Note that p must be in decimal form.
• x: This is the number of “successes.” For this example, we want to find the probability that two of the
children have type-O blood, so x = 2. Note that x must be a whole number.
Putting it all together, type binompdf (5, 0.25, 2) and press enter.
3. Your calculator will return the probability that exactly 2 out of the 5 children have type O blood. You will
find that P(x = 2) = 0.2637. In other words, there is a 26.37% chance that two of the five children will have
type O blood.
12.1 Random digit dialing. When an opinion poll calls residential telephone numbers at random, only 20% of the calls
reach a live person. You watch the random dialing machine make 15 calls. X is the number that reach a live person.
12.6 Random digit dialing. When an opinion poll calls residential telephone numbers at random, only 20%
of the calls reach a live person. You watch the random digit dialing machine make 15 calls.
(a) What is the probability that exactly 3 calls reach a person?
(b) What is the probability that at most 3 calls reach a person?
(c) What is the probability that at least 3 calls reach a person?
(d) What is the probability that fewer than 3 calls reach a person?
(e) What is the probability that more than 3 calls reach a person?
The binomcdf command
Using the question (b) above.
1. Press 2nd VARS for the DISTR menu. Scroll down to binomcdf( and press Enter. This is menu item B if you
have a TI-84 calculator.
2. The syntax for the binomial cumulative density function command is binomcdf(n,p,x).
• n: This is the number of trials. For this example, n = 15 (the number of phone calls).
• p: This is the “success” probability. For this example, p = 0.2 (the probability of reaching a live person).
Note that p must be in decimal form.
• x: This is the number of “successes.” For this example, we want to find the probability that at most 3 calls
reach a person. This is the same as P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3), which is cumulating P(x=0) to
P(x=3). The calculator only uses the left hand side so x = 3 for this example.
Putting it all together, type binomcdf (15, 0.2, 3) and press enter.
3. Your calculator will return the probability that at most 3 calls reach a person. You will find that P(x ≤ 3) =
0.6482. In other words, there is a 64.82% chance that at most 3 calls reach a person.
Binomial mean and standard deviation
mean = expected value = µ = np
Continuing Example 12.3, the count X of children with type O blood is binomial with n = 5 and p = 0.25. The
histogram in Figure 12.1 displays this probability distribution. (Because probabilities are long-run
proportions, using probabilities as the heights of the bars shows what the distribution of X would be in very
many repetitions.) The distribution is skewed to the right.
http://www.youtube.com/watch?v=6d1cKlEfqbQ&feature=channel_video_title