Question

Helen plays basketball. Helen must now attempt two free throws. C = the event that Helen makes the first shot. P(C) = 0.422. D = the event Helen makes the second shot. P(D)= 0.422. The probability that Helen makes the second free throw given that she made the first is 0.516. What is the probability that Helen makes both free throws?

Answer 1

It is given that the event that Helen makes the first shot, P(C) = 0.422 .

The event that Helen makes the second shot, P(D) = 0.422 .

It is also given that the probability that Helen makes the second free throw given that she made the first, P(D|C) = 0.516 .

From conditional probability, we have, P(D|C) = P(D C)/P(C), where event C is given. --------------(1)

From (1), we can say that P(D C) = P(D|C)*P(C) = 0.516*0.422 = 0.2178(rounded up to four decimal places).

Thus, P(D and C) = 0.2178 .

Thus, P(both free throws) = 0.2178 .

Thus, the probability that Helen makes both the free throws = 0.2178 .