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Found in: Page 190

### Introductory Statistics

Book edition OER 2018
Author(s) Barbara Illowsky, Susan Dean
Pages 902 pages
ISBN 9781938168208

# 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 theprobability that Helen makes both free throws?

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

See the step by step solution

## Step 1 : Basic Concepts

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

Pr (A I B) = P (A ∩ B) / P (B)

## Step 2 :

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