S5 January 2006 4 Edexcel

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(ai)

The game on which John wins his second coconut, say X has \mathrm {NB}(2, 0.15) distribution. (since he has probability 0.15 of winning a coconut on any given game) So:

\begin{align*}\mathrm P(X = 7) & = \binom {7 - 1} {2 - 1} (0.15)^2 (1 - 0.15)^5 \\ & = 6 \times 0.15^2 \times 0.85^5 \\ & = 0.0599 \ldots \\ & = 0.0599 \text { (4dp)}\end{align*}

(aii)

The game on which John wins his third coconut, say Y has \mathrm {NB}(3, 0.15) distribution. From the formula booklet we have:

\displaystyle \mathrm E(Y) = \frac r p = \frac 3 {0.15} = 20

(b)

We assume that the probability that John wins a coconut in any given game is constant and that the games are independent events.

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We have:

\displaystyle \mathrm E(G) = \frac rp = 18

and:

\displaystyle \mathrm {var}(G) = \frac {r(1 - p)} {p^2} = 6^2 = 36

So:

r = 18p

and:

r(1 - p) = 36p^2

so:

18p(1 - p) = 36 p^2

since p \ne 0, dividing by 18p gives:

1 - p = 2p

so:

3p = 1

giving:

p = \dfrac 1 3

Since \dfrac 1 3 = 0.333\ldots > 0.15, Sue has a higher probability of winning a coconut in a game.