A fair die is rolled. Consider events E = {1, 3, 5} and  F = {2, 3} then find P((EIF)-

1/2 1/3 1/4 1/6

1/2

When a fair die is rolled, the sample space S will be S= {1, 2, 3, 4, 5, 6} It is given that E = {1, 3, 5} and  F= {2, 3}  So, P(E) = 3/6 = 1/2 P(F) = 2/6 = 1/3 also, E∩F= {3} So P(E∩F) = 1/6 Hence, P(FIF) = P(E∩F) / P(F)                      = (1/6)/(1/3)            P(FIF) = 1/2