Calculates the binomial probability, either at a single value or for all values

for a single value:

binompdf(*trials*, *probability*, *value*

for a list of all values (0 to *trials*)

binompdf(*trials*, *probability*

Press:

- 2ND DISTR to access the distribution menu
- 0 to select binompdf(, or use arrows.

Press ALPHA A instead of 0 on a TI-84+/SE with OS 2.30 or higher.

TI-83/84/+/SE

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This command is used to calculate the binomial probability. In plainer language, it solves a specific type of often-encountered probability problem, that occurs under the following conditions:

- A specific event has only two outcomes, which we will call "success" and "failure"
- This event is going to repeat a specific number of times, or "trials"
- Success or failure is determined randomly with the same probability of success each time the event occurs
- We're interested in the probability that there are exactly N successes

For example, consider a couple that intends to have 4 children. What is the probability that 3 of them are girls?

- The event here is a child being born. It has two outcomes "boy" or "girl". We can call either one a success, but we'll choose to be sexist towards guys and call a girl a success in this problem
- The event is going to repeat 4 times, so we have 4 trials
- The probability of a girl being born is 50% or 1/2 each time
- We're interested in the probability that there are exactly 3 successes (3 girls)

The syntax here is `binompdf(`*trials*, *probability*, *value*). In this case:

`:binompdf(4,.5,3`

This will give .25 when you run it, so there's a .25 (1/4) probability out of 4 children, 3 will be girls.

An alternate syntax for `binompdf(` leaves off the last argument, *value*. This tells the calculator to compute a list of the results for all values. For example:

`:binompdf(4,.5`

This will come to {.0625 .25 .375 .25 .0625} when you run it. These are the probabilities of all 5 outcomes (0 through 4 girls) for 4 children with an equal probability of being born. There's a .0625 probability of no girls, a .25 probability of 1 girl, etc.

# Advanced (for programmers)

The `binompdf(` and `binomcdf(` commands are the only ones apart from `seq(` that can return a list of a given length, and they do it much more quickly. It therefore makes sense, in some situations, to use these commands as substitutes for `seq(`.

Here's how to do it:

`cumSum(``binomcdf(`N,0 gives the list {1 2 … N+1}, and`cumSum(``not(``binompdf(`N,0 gives the list {0 1 2 … N}.- With
`seq(`, you normally do math inside the list: for example,`seq(`3I^{2},I,0,5 - With these commands, you do the same math outside the list: 3
`Ans`^{2}where`Ans`is the list {0 1 … 5}.

An example:

```
:seq(2^I,I,1,5
can be
:cumSum(binomcdf(4,0
:2^Ans
which in turn can be
:2^cumSum(binomcdf(4,0
```

In general (where f() is some operation or even several operations):

```
:seq(f(I),I,1,N
can be
:cumSum(binomcdf(N-1,0
:f(Ans
which can sometimes be
:f(cumSum(binomcdf(N-1,0
```

If the lower bound on I in the `seq(` statement is 0 and not 1, you can use `binompdf(` instead:

```
:seq(f(I),I,0,N
can be
:cumSum(not(binompdf(N,0
:f(Ans
which can sometimes be
:f(cumSum(not(binompdf(N,0
```

This will not work if some command inside `seq(` can take only a number and not a list as an argument. For example, seq(L_{1}(I),I,1,5 cannot be optimized this way.

# Formulas

The value of `binompdf(` is given by the formula

This formula is fairly intuitive. We want to know the probability that out of n trials, exactly k will be successes, so we take the probability of k successes - $p^k$ - multiplied by the probability of (n-k) failures - $(1-p)^{n-k}$ - multiplied by the number of ways to choose which k trials will be successes - $\binom{n}{k}$.

# Error Conditions

**ERR:DOMAIN**is thrown if the number of trials is at least 1 000 000 (unless the other arguments make the problem trivial).**ERR:INVALID DIM**is thrown if you try to generate a list of probabilities with at least 999 trials.

# Related Commands

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