Binomial Probability Distribution on the TI-89

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

See also:  a separate version of these instructions for the TI-83/84

Binomial Probability for One x Value

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

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 = 0.260, x = 2

P(2) = binompdf(4, .260, 2) = 0.2221

Binomial Probability for a Range of x Values

To compute the binomial probability for a range of numbers of successes, use the binomcdf function.

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 = 0.260, 2 ≤ x ≤ 4

P(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

P(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.

P(8 ≤ x ≤ 10) = 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:

n = 24, p = 1/6, 0 ≤ x ≤ 3

P(0 ≤ x ≤ 3) = binomcdf(24, 1/6, 0, 3) = 0.4155