Question

Helen plays basketball. For free throws, she makes the shot 75% of the time. Helen must now attempt two free throws. C = the event that Helen makes the first shot. P(C) = 0.75. D = the event Helen makes the second shot. P(D) = 0.75. The probability that Helen makes the second free throw given that she made the first is 0.85. What is the

probability that Helen makes both free throws?

Short answer

Probability that Helen makes both free throws = 0.6375 or 63.75%

Step-by-step solution

Basic Concepts 

Probability (A given B) = Probability (A & B) / Probability (B)

Pr (A I B) = P (A ∩ B) / P (B)

: 

Pr (C) ie first shot = 0.75 ; Pr (D) ie second shot = 0.75

Pr (D) ie second shot, given Pr (C) ie first shot = 0.85

Pr ( D I C ) = P (C ∩ D) / P (D)

0.85 = P (C ∩ D) / 0.75

P (C ∩ D), ie Pr (C & D) = 0.85 x 0.75 = 0.6375