Question

‘A’ has probability 3/5 of winning a game of chess with ‘B’ If they play four games find the probability that A wins (a) exactly one game (b) atleast 3 games.

Answer 1

Let x be the number of games that ‘A’ wins, is a Binomial variate with parameters n = 4, p =3/5, 

∴ q = 1 – p =2/5

The p.m.f is –

P(x) = nc_xp^xq^{n – x} ; x = 0,1,2, …… n

(a) p (A wins exactly one game)

= 4 × 0.6 × 0.064

= 0.1536.

(b) p (A wins atleast 3 games)

= p(x ≥ 3) = p (x = 3) + p (x = 4)

= 4 × 0.216 × 0.4 + 1 × 0.1296 x 1

= 0.3456 + 0.1296

= 0.4752.