Let X be the number of hit in his 3 bats. Then X ~ Binomial(n = 3, p = 1/3)
A.
Probability that the player gets exactly 1 hit in 3 bats = P(X = 1) = 3C1 * (1/3)1 * (1 - 1/3)3-1
= 3 * (1/3) * (2/3)2
= 0.4444
B.
Probability that the player gets at least 1 hit = P(X 1) = 1 - P(X = 0)
= 1 - 3C0 * (1/3)0 * (1 - 1/3)3-0
= 1 - (2/3)3
= 0.7037
C.
Probability that the player has two hits in his first two at bats = (1/3) * (1/3) = 1/9
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