Ask Question
8 September, 21:59

Suppose a basketball player is practicing shooting, and has a prob-ability. 95 of making each of his shots. Also assume that his shots are in-dependent of one another. Using the Poisson distribution, approximate theprobability that there are at most 2 misses in the first 100 attempts

+3
Answers (1)
  1. 8 September, 22:14
    0
    0.082

    Step-by-step explanation:

    Number of attempts = n = 100

    Since there are only two outcomes and in-dependent of each other, the probability of missing a shot = 1 - Probability of making each shot

    p = 1 - 0.95 = 0.05

    Possion Ratio (λ) = np where n is the number of events and p is the probability of the shot missing

    λ = 100 x 0.05 = 5

    Define X such that X = Number of misses and X ≅ Poisson (λ = 5)

    P [X ≤ 2] = P [X = 0] + P [X = 1] + P [X = 2]

    P [X ≤ 2] = e⁻⁵ + e⁻⁵ x 5 + e⁻⁵ x 5²/2!

    P [X ≤ 2] = e⁻⁵ [1 + 5 + 5²/2!]

    P [X ≤ 2] = e⁻⁵ x 12.25 = 0.082

    The required probability that there are at most 2 misses in the first 100 attempts is 0.082
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “Suppose a basketball player is practicing shooting, and has a prob-ability. 95 of making each of his shots. Also assume that his shots are ...” in 📗 Mathematics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers