Question

Team A and B play 13 games of chess of which 6 are won by A, 4 are won by B and 3 end in a draw. They agree to play a tournament of 4 games. Find The standard deviation of the number of wins by A.

Answer 1

To find the standard deviation of the number of wins by A, we first need to find the probability of A winning each game.

Out of the 13 total games played, A won 6, B won 4, and 3 ended in a draw. This means that the total number of games won, denoted as W, is 6.

The probability of A winning a game is given by:
P(A) = Number of games won by A / Total number of games played = 6 / 13

Since they are playing a tournament of 4 games, the number of games A could potentially win, denoted as X, follows a binomial distribution with parameters n = 4 (number of games in the tournament) and p = 6/13 (probability of A winning each game).

The formula for the standard deviation of a binomial distribution is given by:
σ = sqrt(n * p * (1 - p))

Plugging in the values, we have:
σ = sqrt(4 * (6/13) * (1 - 6/13))
= sqrt(4 * (6/13) * (7/13))
= sqrt((24/13) * (7/13))
= sqrt(168/169)
= sqrt(168) / sqrt(169)
= 12/13

Therefore, the standard deviation of the number of wins by A in the tournament of 4 games is 12/13.