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

for a single value:

binomcdf(*trials*, *probability*, *value*

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

binomcdf(*trials*, *probability*

Press:

- 2ND DISTR to access the distribution menu
- ALPHA A to select binomcdf(, or use arrows.

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

TI-83/84/+/SE

2 bytes

This command is used to calculate the binomial cumulative probability function. 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
**at most**N successes

For example, consider a couple that intends to have 4 children. What is the probability that at most 2 are girls?

- The event here is a child being born. It has two outcomes "boy" or "girl". In this case, since the question is about girls, it's easier to call "girl" a success.
- 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 at most 2 successes (2 girls)

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

`:binomcdf(4,.5,2`

This will give .6875 when you run it, so there's a .6875 probability out of 4 children, at most 2 will be girls.

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

`:binomcdf(4,.5`

This will come to {.0625 .3125 .6875 .9375 1} when you run it. These are all the probabilities we get when you replace "at most 2 girls" with "at most 0", "at most 1", etc. Here, .0625 is the probability of "at most 0" girls (or just 0 girls), .3125 is the probability of at most 1 girl (1 or 0 girls), etc.

Several other probability problems actually are the same as this one. For example, "less than N" girls, just means "at most N-1" girls. "At least N" girls means "at most (total-N)" boys (here we switch our definition of what a success is). "No more than", of course, means the same as "at most".

# 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: seq(3I
^{2},I,0,5 - With these commands, you do the same math outside the list: 3Ans
^{2}where Ans is the list {0 1 … 5}.

```
: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

Since "at most N" is equivalent to "0 or 1 or 2 or 3 or … N", and since we can combine these probabilities by adding them, we can come up with an expression for binomcdf( by adding up values of binompdf(:

(1)(If you're not familiar with sigma notation, $\sum_{i=0}^{k}$ just means "add the following up for each value of i 0 through k")

# Error Conditions

**ERR:DATATYPE**is thrown if you try to generate a list of probabilities with*p*equal to 0 or 1, and at least 257 trials.**ERR:DOMAIN**is thrown if the number of trials is at least 1 000 000, unless the other arguments make the problem trivial.

# Related Commands

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