Question

Australia beat England at cricket 70% of the time when Steve Smith is selected and 10% of the time when he is not (assume no draws). Steve Smith is selected for 80% of test matches (he is suspended the rest of the time). Australia lost the most recent test against England. Given the information above, how likely is it that Steve Smith was selected?

Answer 1

Let A= event that Australia beat England

B=event that Steve Smith is selected.

According to the question, P(A l B) =0.7, P(A l B^c) =0.1,

P(B) =0.8, P(B^c) =1-0.8=0.2.

We have to find the probability that Steve Smith was selected given Australia lost the test, P(B l A^c) .

Now P( AB) =P(A l B)*P(B) =0.7*0.8=0.56 and

P(A^c) =P(A^cB) +P(A^cB^c)

=P(A^c l B) P(B) +P(A^c l B^c) P(B^c)

=[1-P(A l B)]*P(B) +[1-P(A l B^c) ]*P(B^c)

=(1-0.7) *0.8+(1-0.1) *0.2

=(0.3*0.8) +(0.9*0.2)

=0.24+0.18

=0.42

P(A^cB) =P(B) -P(AB) =0.8-0.56=0.24.

Thus, P(B l A^c) =P(A^cB) /P(A^c) =0.24/0.42=0.5714.

Thus, the required probability is 0.5714.