17.1.3 The Geometric Distribution

Geometric Distribution

This describes a game with two outcomes, "win" or "lose." The game is played up to and including the first "win."

📑e.g., win in a driving test.

If the probability of a win is $p$ and $X$ is the number of trials up to and including the first "win," we say $X \sim \text{Geo}(p)$ .

These must be modified to fit the context of the situation:

lightbulbExample

Example: Two is three driving tests, all failures. Find the probability that a person passes on their:

a) 2nd attempt

X \sim \text{Geo}\left(\frac{1}{3}\right)

P(X = 2) = \left(\frac{2}{3}\right) \times \frac{1}{3} = \frac{2}{9}

b) 4th attempt

P(X = 4) = \left(\frac{2}{3}\right)^3 \times \frac{1}{3} = \frac{8}{81}

c) 8th attempt

P(X = 8) = \left(\frac{2}{3}\right)^7 \times \frac{1}{3} = \frac{128}{6561}

infoNote

Summary For a game with two outcomes, win or lose, the probability of the first win occurring on turn $n$ is:

P(X = n) = (1 - p)^{n - 1} p \quad \text{where} \quad X \sim \text{Geo}(p)

lightbulbExample

Example: For $X \sim \text{Geo}(0.7)$ :

$P(X = 8) = (0.3)^7 \times 0.7 = 1.53 \times 10^{-4}$ (7 losses, then a win)

$P(X > 5) = (0.3)^5 = 2.43 \times 10^{-3}$

The bet can be said to be won as soon as the 5th loss has occurred. In the geometric distribution, $>$ probabilities are the easiest to work out.

$0.3^2 = 0.09$

d) $P(X \leq 10) = 1 - P(X > 10)\\$
$= 1 - 0.3^{10}\\$
```
             $= 0.999960451 \approx 1.00 \quad (3 \text{sf})$
```

Do Want: 10, 9, 8

Don't Want: 11,12,13,……..

e) $P(5 \leq X < 7)$ Do Want: 5, 6

                                                                                                 **Don't Want**: 1, 2, 3, 4

$P(X > 4) = P(X > 6) = 0.3^4 - 0.3^6 = 7.37 \times 10^{-3}$