Question

Individuals A and B begin to play a sequence of chess games. Let S be the event that A wins a game, F be the event that B wins a game and suppose that outcomes of successive games are independent with P(S) = p and P(F) = 1 − p (they never draw). They will play until one of them wins ten games. Let X denote the number of games played (with possible values 10, 11, ... , 19). For x = 10, 11, ... , 19, obtain an expression for the probability mass function for X. (8 marks)

Answer 1

The probability mass function (PMF) for X, the number of games played, is
`P(x) = \left\{ \begin{array}{lr} p^(x-1)(1-p), & \text{for } x = 10, 11, \ldots, 19 \\ 0, & \text{Otherwise } \end{array}`

How to determine the expression for the probability mass function

From the question, we have the following parameters that can be used in our computation:

• p is the probability that A wins a game
• 1 - p is the probability that B wins a game

This means that the probability mass function (PMF) for X is

`P(x) = \left\{ \begin{array}{lr} p^(x-1)(1-p), & \text{for } x = 10, 11, \ldots, 19 \\ 0, & \text{Otherwise } \end{array}`

Where

• P(x) represents the probability mass function for the random variable x, which indicates the number of games played.
• pˣ ⁻ ¹ represents the probability of A winning in a single game, given that there are x - 1 games before the x-th game
• (1 - p) represents the probability of B winning in a single game

Lastly, we consider cases for x ranging from 10 to 19, as the game must end by the 19th round.

And if x is not in the range 10 to 19, the probability is zero.