Question

Helen plays basketball. For free throws, she makes the shot 78% of the time. Helen must now attempt two free throws. C = the event that Helen makes the first shot. P(C) = 0.78. D = the event Helen makes the second shot. P(D) = 0.78. The probability that Helen makes the second free throw given that she made the first is 0.86. What is the probability that Helen makes both free throws?

Answer 1

Answer:

0.6708 or 67.08%

Step-by-step explanation:

Helen can only make both free throws if she makes the first. The probability that she makes the first free throw is P(C) = 0.78, now given that she has already made the first one, the probability that she makes the second is P(D|C) = 0.86. Therefore, the probability of Helen making both free throws is:

$P(C+D) = P(C) *P(D|C) = 0.78*0.86\\P(C+D) = 0.6708$

There is a 0.6708 probability that Helen makes both free throws.