TC3 → Stan Brown → TI-83/84/89 → Binomial PD (TI-83/84)
revised 17 Sep 2011

Binomial Probability Distribution on TI-83/84

Copyright © 2007–2014 by Stan Brown, Oak Road Systems

Summary: With your TI-83/84, you can do all types of probability calculations for a binomial probability distribution. (This page is for students who don’t have MATH200A Program part 3.)

See also: TI-89/92 users can use the procedure in Binomial Probability Distribution on TI-89 to compute binomial probability.

Binomial Probability for One x Value

To compute the binomial probability for one particular number of successes, use the binompdf function.

Keystrokes [2nd VARS makes DISTR] [ 6 times] [STO→] [2nd 3 makes L3]
Format binompdf(n, p, x) → list

Example 1: Larry’s batting average is .260. If he’s at bat four times, what is the probability that he gets exactly two hits?

Solution:

n = 4, p = 0.26, x = 2

Note: Some textbooks use r for number of successes, rather than x.

binompdf(4,.26,2) = 0.2221

Binomial Probability for a Range of x Values

To compute the binomial probability for a range of numbers of successes from xlow to xhigh, use the binompdf function again, but store its value into a statistics list and then use sum to add up the necessary probabilities from the list. You can use any list; I’ll use L3.

Keystrokes [2nd VARS makes DISTR] [ 6 times] [STO→] [2nd 3 makes L3]
Format binompdf(n, p, x) → list
Keystrokes [2nd STAT makes LIST] [] [5]
Format sum(list, xlow+1, xhigh+1)

Example 2: Larry’s batting average is .260. If he’s at bat six times, what is the probability that he gets two to four hits?

Solution:

n = 6, p = 0.26, 2 ≤ x ≤ 4

binompdf(6,.26) → L3

sum(L3,3,5) = 0.4840

Remember: add 1 to xlow and xhigh. To get the probability of 2 to 4 successes, you have to put the numbers 3 and 5 in the sum function.

Example 3: Suppose 65% of the registered voters in Dryden are Republicans. In a random sample of ten registered voters, what’s the probability of fewer than six Republicans?

Solution: “Fewer than six” is zero through five.

n = 10, p = 0.65, 0 ≤ x ≤ 5

binompdf(10,.65) → L3

sum(L3,1,6) = 0.2485

There’s about one chance in four of getting fewer than six Republicans in a random sample of ten registered voters.

Example 4: With the same data, what’s the probability of getting eight or more Republicans?

Solution: “Eight or more” means eight to ten since there are only ten trials.

n = 10, p = 0.65, 8 ≤ x ≤ 10

You don’t have to repeat binompddf because the values are still in L3.

sum(L3,9,11) = 0.2616

Remember: add 1 to xlow and xhigh. To get the probability of 8 to 10 successes, you have to put the numbers 9 and 11 in the sum function.

Example 5: A fair die has a 1/6 chance of rolling a 2. In 24 rolls, what's the probability of getting no more than three 2’s?

Solution: “No more than three” means zero to three.

n = 24, p = 1/6, 0 ≤ x ≤ 3

binomcdf(24,1/6) → L3

sum(L3,1,4)= 0.4155


This page is used in instruction at Tompkins Cortland Community College in Dryden, New York; it’s not an official statement of the College. Please visit www.tc3.edu/instruct/sbrown/ to report errors or ask to copy it.

For updates and new info, go to http://www.tc3.edu/instruct/sbrown/ti83/