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Chapter 4: Problem 4

A particular basketball player hits \(70 \%\) of her free throws. When she tosses a pair of free throws, the four possible simple events and three of their associated probabilities are as given in the table: $$\begin{array}{cllc} & \text { Outcome } & \text { Outcome } & \\\\\text { Simple } & \text { of First } & \text { of Second } & \\\\\text { Event } & \text { Free Throw } & \text { Free Throw } & \text { Probability } \\\\\hline 1 & \text { Hit } & \text { Hit } & .49 \\\2 & \text { Hit } & \text { Miss } & ? \\\3 & \text { Miss } & \text { Hit } & .21 \\\4 & \text { Miss } & \text { Miss } & .09\end{array}$$ a. Find the probability that the player will hit on the first throw and miss on the second. b. Find the probability that the player will hit on at least one of the two free throws.

Short Answer

Expert verified
Answer: The probability that the player will hit on the first throw and miss on the second is 0.21. The probability that the player will hit at least one of the two free throws is 0.91.

Step by step solution

01

Calculate the probability of hitting the first throw and missing the second

Since we know the probability of hitting a free throw is \(70 \%\) and missing it is \(30 \%\), we can multiply these probabilities to find the probability of hitting the first throw and missing the second. So, \(P(\)Hit first, Miss second\() = P(\)Hit\() * P(\)Miss\()\) \(= 0.7 * 0.3 = 0.21\)
02

Find the probability of hitting at least one of the two free throws

To find the probability of hitting at least one of the two free throws, we can simply add the probabilities of the events where she hits at least one: Event 1: Hit both free throws - Probability \(= 0.49\) Event 2: Hit the first and miss the second - Probability \(= 0.21\) Event 3: Miss the first and hit the second - Probability \(= 0.21\) \(P(\)Hit at least one\() = P(\)Event 1\() + P(\)Event 2\() + P(\)Event 3\()\) \(= 0.49 + 0.21 + 0.21 = 0.91\) So, a. The probability that the player will hit on the first throw and miss on the second is \(0.21\). b. The probability that the player will hit at least one of the two free throws is \(0.91\).

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