Question

Fischer and Spassky play a chess match in which the first player to win a game wins the match. After 10 successive draws, the match is declared drawn. Each game is won by Fischer with probability 0.4, is won by Spassky with probability 0.3, and is a draw with the probability 0.3 independent of previous games.
(a) What is the probability that Fischer wins the match?
(b) What is the PMF of the duration of the match?

Answer 1

Answer:

a)

the probability that Fischer wins the match is 0.5714

b)

p( D = d ) = { $( 0.3 )^{d-1$ $(0.7)$,  d = 1, 2, 3, ........

                                 $0$                     otherwise

Step-by-step explanation:

Given that;

probability that Fischer wins the match, p = 4

probability that Spassky wins the match, q = 0.3

Match drawn, 1 - p - q = 1 - 0.4 - 0.3 = 0.7

(a) What is the probability that Fischer wins the match?

P( Fischer wins) = p/( p+q)

we substitute

P( Fischer wins) = 0.4 / ( 0.4 + 0.3)

P( Fischer wins) = 0.4 / 0.7

P( Fischer wins) = 0.5714

Therefore, the probability that Fischer wins the match is 0.5714

b) What is the PMF of the duration of the match?

let D represent the duration of the match

since the duration D of the match is a geometric random variable with parameter p + q,

the PMF will be;

p( D = d ) = $( 1 - p - q )^{d-1$ ( p + q )

= $( 1 - 0.4 - 0.3 )^{d-1$ ( 0.4 + 0.3 )

p( D = d ) = $( 0.3 )^{d-1$ ( 0.7)

that is, p( D = d ) = { $( 0.3 )^{d-1$ $(0.7)$,  d = 1, 2, 3, ........

                                 $0$                     otherwise