Justin is a basketball player. The probability that he makes a half-court shot is 0.04. Suppose Justin shoots 130 half-court shots.(a)What is the probability that Justin makes exactly 6 of these half-court shots? Calculate this probability using a binomial distribution. Round your answers to four decimal places.(b)Use the Poisson distribution to approximate the probability that Justin makes exactly 6 of these half-court shots. Round your answer to four decimal places.
Question
Justin is a basketball player. The probability that he makes a half-court shot is 0.04. Suppose Justin shoots 130 half-court shots.(a)What is the probability that Justin makes exactly 6 of these half-court shots? Calculate this probability using a binomial distribution. Round your answers to four decimal places.(b)Use the Poisson distribution to approximate the probability that Justin makes exactly 6 of these half-court shots. Round your answer to four decimal places.
Solution
(a) To calculate the probability of Justin making exactly 6 half-court shots using a binomial distribution, we use the formula:
P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k))
where:
- P(X=k) is the probability of k successes in n trials
- C(n, k) is the combination of n items taken k at a time
- p is the probability of success (in this case, making a shot)
- n is the number of trials (in this case, the number of shots)
- k is the number of successes we're interested in (in this case, 6 shots)
So, plugging in the given values:
P(X=6) = C(130, 6) * (0.04^6) * ((1-0.04)^(130-6))
Calculating this gives us a probability of approximately 0.1857.
(b) To approximate the probability using a Poisson distribution, we use the formula:
P(X=k) = (λ^k * e^-λ) / k!
where:
- λ is the expected number of successes, which is np (in this case, 1300.04 = 5.2)
- e is the base of the natural logarithm (approximately 2.71828)
- k is the number of successes we're interested in (in this case, 6)
So, plugging in the given values:
P(X=6) = (5.2^6 * e^-5.2) / 6!
Calculating this gives us a probability of approximately 0.1497.
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