*Chapter P, Section 1, Exercise 037 Free Throws During the 2015-2016 NBA season,

Question

*Chapter P, Section 1, Exercise 037 Free Throws During the 2015-2016 NBA season, Steven Adams of the Oklahoma City Thunder had a free throw shooting percentage of 0.502. Assume that the probability Steven Adams makes any given free throw is fixed at 0.502, and that free throws are independent. Let S denote successfully making a free throw and F denote missing it. (a) If Steven Adams shoots two free throws, what is the probability that he makes both of them? Round your answer to three decimal places. P(Makes two) - Answer1: the absolute tolerance is +-0.005 (b) If Steven Adams shoots two free throws, what is the probability that he misses both of them? Round your answer to three decimal places. P(Misses two)- Answer *1: the absolute tolerance is -0.005 (c) If Steven Adams shoots two free throws, what is the probability that he makes exactly one of them? Round your answer to three decimal places. P(Makes exactly one)- Answer1: the absolute tolerance is +-0.005

Explanation / Answer

X ~ Binomial (n,p)

Binomial probability distribution is

P(X) = nCx px ( 1 - p)n-x

a)

p = 0.502 , 1 - p = 1 - 0.502 = 0.498 , n = 2

P( makes 2) = P(X = 2)

= 2C2 0.5022 0.4980

= 0.252

b)

P( misses 2) = P( X = 0) = probability of none makes

= 2C0 0.5020 0.4982

= 0.248

c)

P(Makes exactly one) = P(Makes first) * P(Second miss) + P(First miss) * P(Second makes)

= 0.502 * 0.498 + 0.498 * 0.502

= 0.500

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.