Two important members of a cricket team are injured, and each has probability 1/3 of recovering before the match. The recoveries of the two players are independent of each other. If both are able to play then the team has probability 3/4 of winning the match, if only one of them plays then the probability of winning is 1/2 and if neither play the probability of winning is 1/16. What is the probability that the match is won?
The following is what i have done:
This is total probablity and we need to find the probabilty that the match is won.
So A= event one players plays
B= event both players play
C= event nither plauers play
Let D be the event that they win the match.
P(D)= P(D l A) * P(A) + P(D l B) * P(B) + P(D l C) * P(C)
P(D l A) = 1/2 P(A)= 1/3
P(D l B) = 3/4 P(B)= 1/3 * 1/3 = 1/9
P(D l C) = 1/16 P(C) = (1-1/3)^2 = 4/9
so i got my answers as 1/2 * 1/3 + 3/4 * 2/3 + 1/16 * 4/9 = 5/18
Can Someone Please Cheak If My Answer Is Correct As I Am Not Really Sure.