TC3 → Stan Brown → TI-83/84/89 → Binomial PD (TI-89/92)
revised 17 Sep 2011

Binomial Probability Distribution on TI-89

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

Summary: With your TI-89/92, you can do all types of probability calculations for a binomial probability distribution.

See also: TI-83/84 users can use the program in MATH200A Program part 3 or the calculator procedure in Binomial Probability Distribution on TI-83/84 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.

TI-89 Home ScreenTI-89 Stats/List Editor
Keystrokes [CATALOG] [F3] [plain ( makes B] [] [F5] [ALPHA ( makes B]
Format binompdf(n, p, x) (dialog box)

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 binomcdf function.

TI-89 Home ScreenTI-89 Stats/List
Keystrokes [CATALOG] [F3] [plain ( makes B] [F5] [ALPHA ) makes C]
Format binomcdf(n, p, xlow, xhigh) (dialog box)

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

binomcdf(6,.26,2,4) = 0.4840

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

binomcdf(10, .65, 0, 5) = 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.

binomcdf(10,.65,8,10) = 0.2616

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(24,1/6,0,3) = 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/