Question

Three boys play a game of luck in which their respective chances of winning are 1/2 1/3 and 1/4.What is the probability that one and only one of the boys wins the game?

Answer 1

Pr(first boy) = A = 1/2
Pr(second boy) = B = 1/3
Pr(third boy) = C = 1/4

A, B and C can win = 1/2*1/3*1/4
=1/24
A wins, B wins, C looses=1/2*1/3*3/4
3/24
A wins, B looses, C wins=1/2*2/3*1/4
2/24
A looses, B wins , C wins=1/2*1/3*1/4
1/24
Probability at least one of them win=1/24 +3/24 + 2/24 + 1/24
=1/24 +3/24 + 2/24 + 1/24
=7/24