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Ted Williams has a lifetime batting average of 0.344. Assume that Ted Williams came to bat officially four times every game played. What would be Ted Williams's probability of getting at least two hits in a game?

User AlexGad
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Final answer:

To calculate Ted Williams's probability of getting at least two hits in a game, we need to use the binomial distribution formula. The binomial distribution formula is P(X ≥ k) = 1 - P(X < k), where X follows a binomial distribution with parameters n, the number of trials, and p, the probability of success on each trial. In this case, n = 4 and p = 0.344. We need to find P(X ≥ 2), which is equal to 1 - P(X < 2).

Step-by-step explanation:

To calculate Ted Williams's probability of getting at least two hits in a game, we need to use the binomial distribution formula. The binomial distribution formula is P(X ≥ k) = 1 - P(X < k), where X follows a binomial distribution with parameters n, the number of trials, and p, the probability of success on each trial. In this case, n = 4 and p = 0.344. We need to find P(X ≥ 2), which is equal to 1 - P(X < 2). Using the binomial distribution formula, we can calculate P(X < 2) as follows:

Calculate P(X = 0): P(X = 0) = (4 choose 0) * (0.344^0) * (1 - 0.344)^(4-0) = 0.442077824

Calculate P(X = 1): P(X = 1) = (4 choose 1) * (0.344^1) * (1 - 0.344)^(4-1) = 0.40906009

Calculate P(X < 2) = P(X = 0) + P(X = 1) = 0.442077824 + 0.409060091 = 0.851137915

Calculate P(X ≥ 2) = 1 - P(X < 2) = 1 - 0.851137915 = 0.148862085

Therefore, Ted Williams's probability of getting at least two hits in a game is approximately 0.1489.

User Evgeny Shadchnev
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