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3.15

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Introductory Statistics

Found in: Page 190

Book edition OER 2018

Author(s) Barbara Illowsky, Susan Dean

Pages 902 pages

ISBN 9781938168208

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Short Answer

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?

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

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

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

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