Question

It has been found that if A and B play a game 12 times, A wins 6 times, B wins 4 times and they draw twice. A and B take part in a series of 3 games. The probability that they will win alternately is
A. `(5)/(72)`
B. `(5)/(36)`
C. `(19)/(27)`
D. none of these

Answer 1

Correct Answer - B
Let E be the event that A wins the game and F be the event that B wins the game. Then,
`P(E )=(6)/(12)=(1)/(2) " and " P(F)=(4)/(12)=(1)/(3)`
`therefore` Required probability `=P{(E cap F cap E) cup (F cap E cap F)}`
`=P(E cap F cap E)+P(F cap E cap F)`
`=P(E )P(F) P(E )+P(F) P(E )P(F)`
`=(1)/(2)xx(1)/(3)xx(1)/(2)+(1)/(3)xx(1)/(2)+(1)/(3)xx(1)/(2)xx(1)/(3)=(5)/(36)`