Binomial Probability Distribution on the TI-83/84

Summary:  With your TI-83 or TI-84, you can do all types of probability calculations for a binomial probability distribution. This note includes a downloadable TI-83 program, but even without the program you can compute these probabilities using native TI-83/84 functions.

TI-83/84 Program BINOMPRB

The TI-83 and TI-84 can compute binomial probabilities for a single number of successes or a range of numbers of successes, but the details are a bit finicky. I’ve written a program to simplify this for you.

Getting the Program

There are three methods to get the program into your calculator:

• If a classmate has the program on her calculator (any model TI-83/84), she can transfer it to yours with the short cable with headphone-style plugs that comes with all TI calculators. On your calculator, press [2nd x,T,θ,n makes LINK] [►] [ENTER], and then on hers press [2nd x,T,θ,n makes LINK] [3], select the program, then press [►] [ENTER].
• Or, download BINOMPRB.ZIP (18 KB, updated Jun 21, 2008), unzip it, and transfer file BINOMPRB.8XP to your calculator. (This requires TI-Connect or TI-GraphLink software and a cable.)
• Or, as a last resort, key in the program. See BINOMPRB.PDF and BINOMPRB_HINTS.HTM in the BINOMPRB.ZIP file.

Using the Program

Press [PROG], scroll to BINOMPRB, and press [ENTER] twice. The program will prompt you for the sample size or number of trials (N), the probability of success on any one trial (P), and the desired numbers of successes (FROM and TO, which will be the same number if you want one particular number of successes).

The program will validate your inputs, and will insist that you correct anything invalid. If you want to break out of this or any TI-83/84 program, press [ON] and then [1] at the menu.

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

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

Here you want the probability of exactly two successes, so FROM and TO will both be 2. In other words, you are computing the probability of 2 through two successes. Run the BINOMPRB program with

N = 4, P = .26, FROM = 2, TO = 2.

The input and output screens are shown at right.

Answer: P(2) = 0.2221

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

BINOMPRB program: N = 6, P = .26, FROM = 2, TO = 4

Answer: 0.4840

The output screen is shown at right.

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

BINOMPRB program: N = 10, P = .65, FROM = 0, TO = 5

Answer: 0.2485

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

Behind the Scenes

How does the program work? You might like to examine the source code, which is in PDF form in the download file BINOMPRB.ZIP.

As with most programs, the bulk of the code is getting and validating your input: are N, FROM, and TO integers, is P between 0 and 1, are FROM and TO less than N, is FROM ≤ TO. (TI-83/84 variables are limited to one letter, so the program uses X for FROM and Y for TO.)

The next largest chunk is the display of results: to get everything on one screen the program must display all the numbers first, and then go back and overlay them with the vaiable names.

The actual calculations are fairly short. The program uses the native TI-83/84 functions binompdf and binomcdf, which compute the binomial probability of X number of successes or of 0 through X successes.

If you enter FROM and TO as the same number, the answer is easy: binompdf(N,P,FROM).

If TO > FROM, things are a bit trickier. One possibility would be a loop, adding up binompdf(N,P,FROM) through binompdf(N,P,TO). But a shorter calculation is possible. The probability of, say 8 through 11 successes (8, 9, 10, 11) is the same as the probability of 0 through 11 successes, minus the probability of 0 through 7 successes. More generally, the probability of FROM through TO successes is the probability of 0 through TO successes, minus the probability of FROM−1 successes. And that’s the calculation the program does. As a special case, if FROM = 0 there’s no subtraction because the number of successes can’t be less than zero.

Variables N, P, X (FROM), Y (TO), and Z (answer) are all left for you to examine if you wish. Press [ALPHA LOG makes N] or another [ALPHA] combination to use any of them in a calculation. If you want to recover the small amount of RAM they use, press [2nd + makes MEM] [2] [2] and then use the [DEL] key.

TI-83/84 Native Functions

If you don’t have the BINOMPRB program, you can still compute probabilities of the binomial distribution by using the TI-83 or TI-84’s native functions binompdf and binomcdf.

Binomial Probability for One x Value

To compute the binomial probability of 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

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

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

Binomial Probability for any Range of x Values

To compute the binomial probability for a range of numbers of successes, you need two steps: use the binompdf function to store the whole distribution in a statistics list, then use the sum function to add up the probabilities in the relevant part of the list.

Note that you must add 1 to the xlow and xhigh numbers. Why? because the x’s start at 0 but the row numbers in the list start at 1. To see this, go into the Stat Edit screen after storing the distribution in your list.

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

binompdf(6, .26) → L3

P(2≤x≤4) = sum(L3, 3, 5) = 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

binompdf(10, .65) → L3

P( x < 6 ) = P( x ≤ 5 ) = 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

P( x ≥ 8 ) = sum(L3, 9, 11) = 0.2616

Optional Shortcut for P(x ≤ a number)

To compute the probability of any number of successes up to a certain number, you can use the above procedure or you can use binomcdf and skip the sum function.

Use the actual value of xhigh with binomcdf — don’t add 1.

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

Optional Shortcuts for P(x < number), P(x ≥ number), P(x > number)

binomcdf is your friend for all types of problems for a range of numbers of successes where the range starts at 0 or ends at n. Just think about the inequality in each problem, and change it to ≤ form. You may also need to use the complement.

• P(x < 5) is the same as P(x ≤ 4).
  In words, less than five means the same as four or less (for whole numbers).
• P(x > 5) is the same as 1−P(x ≤ 5).
  In words, you can either have more than five successes, or you can have five successes or less.
• P(x ≥ 5) is the same as P(x > 4), which is the same as 1−P(x ≤4).
  In words, you can either have five or more successes, or you can have four successes or less.

Here’s how some previous examples work with binomcdf:

Example 3 redone: 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 less than or equal to five.

n = 10, p = 0.65, x ≤ 5

P( x < 6 ) = P( x ≤ 5 ) = binomcdf(10, 0.65, 5) = 0.2485

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

Solution: “Eight or more” is the complement of “seven or less”. binomcdf gives the complement of the desired probability:

n = 10, p = 0.65, x ≥ 8

P( x ≥ 8 ) = 1 − P( x ≤ 7 ) = 1−binomcdf(10, 0.65, 7) = 0.2616