A high school baseball player has a 0.181 batting average. In one game, he gets 8 at bats. What is the probability he will get at least 5 hits in the game?
To find the probability that the baseball player will get at least 5 hits in the game, we can use the binomial distribution, since the player either gets a hit or does not get a hit on each at bat, and the at bats are independent.
Let X be the number of hits the player gets in the game. Then X follows a binomial distribution with parameters n = 8 (the number of at bats) and p = 0.181 (the probability of getting a hit on each at bat).
The probability of getting at least 5 hits is equal to the sum of the probabilities of getting 5, 6, 7, or 8 hits:
P(X ≥ 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)
Using the formula for the probability mass function of the binomial distribution, we can calculate each of these probabilities:
P(X = k) = (n choose k) p^k (1-p)^(n-k)
where (n choose k) is the binomial coefficient.
Therefore, we have:
P(X = 5) = (8 choose 5) (0.181)^5 (1-0.181)^(8-5) ≈ 0.222
P(X = 6) = (8 choose 6) (0.181)^6 (1-0.181)^(8-6) ≈ 0.079
P(X = 7) = (8 choose 7) (0.181)^7 (1-0.181)^(8-7) ≈ 0.013
P(X = 8) = (8 choose 8) (0.181)^8 (1-0.181)^(8-8) ≈ 0.001
Therefore, the probability of getting at least 5 hits in the game is:
P(X ≥ 5) = 0.222 + 0.079 + 0.013 + 0.001 = 0.315
Therefore, the probability that the baseball player will get at least 5 hits in the game is approximately 0.315 or 31.5% (rounded to one decimal place).
