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

Answer:0.082Step-by-step explanation:Number of attempts = n = 100Since there are only two outcomes and in-dependent of each other, the probability of missing a shot  = 1 - Probability of making each shotp = 1 - 0.95 = 0.05Possion Ratio (λ) = np where n is the number of events and p is the probability of the shot missingλ = 100 x 0.05 = 5Define 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 = 0.082The required probability that there are at most 2 misses in the first 100 attempts is 0.082