Another TI-BASIC challenge proposed to me by the same person who asked me last time was to calculate the probability mass function of a multinomial distribution. The formula is:

## How it works

This formula can be broken down into three basic parts:

For each $i$, $x_i$ and $p_i$ are read, and the new values of $numerator$ and $denominator$ are calculated. Once this is done, $x_i$ and $p_i$ are no longer needed. Finally, once all of the numbers have been read in, the final answer is just

This program also does some sanity checking to make sure that $\sum_{i=1}^k p_i = 1$ and $\sum_{i=1}^k x_i = n$.

## Variables used

`N` | $n$ |

`D` | $denominator$ ($x_1 \cdots x_k$) |

`M` | $multiplier$ ($p_1^{x_1}\cdots p_k^{x_k}$) |

`P` | $p\_i$ |

`O` | $\sum_{i=1}^k p_i$ |

`X` | $x\_i$ |

`W` | $\sum_{i=1}^k x_i$ |

## The Program

Download: MULTINOM.8Xp

```
Prompt N
1→D:1→M:0→O:0→W:1→X
While X≠0
Prompt X
If X≠0:Then
Prompt P
X+W→W:P+O→O
D*(X!)→D
M*(P^X)→M
End:End
If O≠1:Then
Disp "E: sum(P)≠1
Stop:End
If W≠N:Then
Disp "E: sum(X)≠N
Stop:End
((N!)/D)*M
```

## Example

In a recent three-way election for a large country, candidate A received 20% of the votes, candidate B received 30% of the votes, and candidate C received 50% of the votes. If six voters are selected randomly, what is the probability that there will be exactly one supporter for candidate A, two supporters for candidate B and three supporters for candidate C in the sample? (source)

To find the solution using `prgmMULTINOM`

:

`N=?` | Number of supporters (1+2+3) |

`X=?` | Supporters for candidate A |

`P=?` | Votes for candidate A |

`X=?` | Supporters for candidate B |

`P=?` | Votes for candidate B |

`X=?` | Supporters for candidate C |

`P=?` | Votes for candidate C |

`X=?` | End of input |

`.135` | Result |

Update December 2015: Added example of how to use the program, since I couldn’t remember how myself.