Question

A,B, and C play a game and chances of their winning it in an attempt are 2/3,1/2 and 1/4 respectively. A has the first chance, followed by B and then by C. This cycle is repeated till one of them wins the game. Find their respective chances of winning the game.

Answer 1

hєчα mαtє

In the 1/3 of games in which A doesn't win, B has a 1/2 chance of winning in the second game (for a total chance of 1/2*1/3 = 1/6 of winning). So with probability 1- (2/3 + 1/6) = 1 - 5/6 = 1/6 we reach the third game, where C has a 1/4 chance of winning, for a total chance of C winning of 1/4*1/6 or 1/24.

plzz mαrk αѕ вrαnlíѕt

Answer 2

777+48858%

78&&%#-_4457576

$5222123 \times \frac{?}{?}$

gkggihhdir