The negative binomial distribution is a Discrete Distribution of Random Variables

- Geometric Distribution: probability of observing first success on $n$th trial
- NBD: probability of observing $k$th success on $n$th trial
- so NBD is a generic case of Geometric Distribution

A distribution is NBD if:

- trials are independent
- each trial is a Bernoulli Trial - i.e. has only two outcomes - success and failure
- $p$ is the same for all the trials
- the last trial must be success

NBD:

- $k$ - number of successes, $n$ - total number of trials
- $p$ - probability of success, $q = 1 - p$ - probability of failure
- pmf: $Pr(X = x) = C^{k-1}_{n-1} q^{n-k} p^{k}$

A footballer can go home only after he scores 4th goal

- $p$ - probability of success

Suppose he made 6 attempts

- what's the probability that he scored 4 goals, and
__the last trial led to success__?

Let's write down all possible sequences when the 4th kick is on the 6th attempt:

from itertools import permutations p = set([x for x in permutations('SSSSFF')]) [x for x in p if x[-1] == 'S']

1 | F | S | S | F | S | S |
---|---|---|---|---|---|---|

2 | S | S | F | S | F | S |

3 | S | F | F | S | S | S |

4 | F | F | S | S | S | S |

5 | F | S | F | S | S | S |

6 | S | F | S | S | F | S |

7 | S | F | S | F | S | S |

8 | S | S | F | F | S | S |

9 | S | S | S | F | F | S |

10 | F | S | S | S | F | S |

There are 10 sequences that lead to this outcome

- note that Success is always last!

Let's calculate the probability of going home after 6 kicks (having 6th kick successful)

- so $P(\text{go home}) = \sum_{i=1}^{10} P(\text{seq}_i)$
- each sequence has the same probability of occurring:
- $P(\text{seq}_i) = P(\text{seq}) = q^{n-k} p^{k}$
- this is the probability of observing $n-k$ failures and $k$ successes

- there are ${n - 1 \choose k - 1}$ ways to pick these elements ($C_{n - 1}^{k - 1}$)