Question

. Two boys A and B throw a ball at a target. Suppose that the probability that boy A will hit the target on any throw is 1/3 and the probability that boy B will hit the target on any throw is 1/4. Suppose also that boy A throws first and the two boys take turns throwing. Determine the probability that the target will be hit for the first time on the third throw of boy A.

Answer 1

Solution:-

P(A)=1/3

P(B)=1/4

Probability that boy A will not hit the target in first attempt=P(A')=(1-1/3)=2/3

Probability that boy A will not hit the target in first attempt=P(b')=(1-1/4)=3/4

So, A throws and misses, B throws and misses.

So, A throws and misses for 2nd time, B throws and misses for the 2nd time And then A throws and gets it.

Probability that the target will be hit for the first time on the third throw of boy A=(2/3)*(3/4)*(2/3)*(3/4)*(1/3)

=0.0833