# Binomial Probability Distribution on TI-89

Copyright © 2007–2016 by Stan Brown

Copyright © 2007–2016 by Stan Brown

**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 part 3
or the calculator procedure
here,
in Stats without Tears,
to compute binomial probability.

To compute the binomial probability for
**one particular number of successes**, use the `binompdf`

function.

TI-89 Home Screen | TI-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

To compute the binomial probability for a
**range of numbers of successes** from *xlow* to
*xhigh*, use the
`binomcdf`

function.

TI-89 Home Screen | TI-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

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