- #1

Gridvvk

- 56

- 1

## Homework Statement

This isn't really homework, just reviewing for a test. This is problem 3.17 in 'A modern introduction to probability and statistics: understanding why and how' Dekking.

But since it can be seen as a HW problem, might as well post here.

Question:

You and I play a tennis match. It is deuce, which means if you win the next two rallies, you win the game; if I win both rallies, I win the game; if we each win one rally, it is deuce again. Suppose the outcome of a rally is independent of other rallies, and you win a rally with probability p. Let W be the event you win the game, G the game ends after the next two rallies, and D it becomes deuce again.

(a) Determine P(W|G).

(b) Show that P(W) = p^2 + 2p(1 - p)P(W|D) and use P(W) = P(W|D) (why is this so?)

to determine P(W).

(c) Explain why the answers are the same.

**The attempt at a solution**

(a) P(W|G) = p^2

Since you have a probability p of winning each rally (and they are independent).

(b) P(W) = P(W ∩ G) + P(W ∩ D) , since D and G are mutually exclusive exhaustive events

P(W) = P(W|G)P(G) + P(W|D)P(D) = p^2 * P(G) + p(1-p) P(W|D)

I know P(W|G) = p^2 from (a), and P(D) = p(1-p) since we would each have to win one rally for it go to deuce.

I am not sure how to compute P(G), I thought it should P(G) = p^2 + (1-p)^2, either you win both or I win both; however, this doesn't give the desired result what we were supposed to show.

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P(W) = P(W|D) : I suppose because if it's a Deuce we "reset" and have an equal chance of winning again.

Taking what we had to show for granted I can solve for P(W)

P(W) = p^2 + 2p(1-p)P(W|D)

P(W) - 2p(1-p)P(W) = p^2

P(W)(1 - 2p(1-p)) = p^2

P(W) = p^2 / (1 - 2p + p^2)

(c) What two answers is the question referencing to being the same?

Was I supposed to get the same answer for P(W) and P(W|G)?