The geometric distribution

Three quick questions from earlier lessons.

You roll a die until the first six. What is the probability that the first six comes on roll 3?

Waiting for the first success.

Geometric setting: independent trials with the same

p

, repeated until the first success;

X

= the number of the trial of the first success.

P(X = k) = (1 - p)^{k - 1}p

,

\mu_X = \dfrac{1}{p}

.

$p = 0.25$ .

P(X = 2) = 0.75 \times 0.25 = 0.1875

First success on trial 2.

P(X \le 2) = 1 - 0.75^2 = 0.4375

Within 2 trials: complement of 2 failures.

\mu = 4

1/p.

Your turn: $p = 0.1$ . The mean number of trials.

10

1/0.1.

Each call reaches a customer with probability 1/2, independently. Find P(the first success is on call 2).
Each call reaches a customer with probability 1/4, independently. Find P(the first success is on call 2).

A die-like spinner shows 'win' with probability 1/10. On average, how many spins are needed to get the first win?

Each attempt succeeds with probability 1/7. Find P(the first success comes within the first two attempts).
Each attempt succeeds with probability 1/9. Find P(the first success comes within the first two attempts).

Three questions to check you have got it.

Each call reaches a customer with probability 1/2, independently. Find P(the first success is on call 2).
Each call reaches a customer with probability 1/4, independently. Find P(the first success is on call 2).
Each call reaches a customer with probability 1/2, independently. Find P(the first success is on call 6).

Geometric distribution

$P(X = k) = (1 - p)^{k - 1}p$ , $\mu = 1/p$ .

Watch out: $k - 1$ failures before the first success.

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