# Binomial Probability Distribution on TI-83/84

Copyright © 2007–2015 by Stan Brown, Oak Road Systems

Copyright © 2007–2015 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.

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

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/