A baseball player has a .300 batting average; (he gets a hit in 30% of his total

Question

A baseball player has a .300 batting average; (he gets a hit in 30% of his total “at-bats” (opportunities)). In a particular week, he will have a total of 20 at-bats. Assume that two outcomes are possible at each at-bat, a hit or an “out”. Also, assume that the probability of a hit is constant and that the at-bats are independent.

a. What is the expected number of hits for the week?

b. What is the probability that he gets more than six hits (hits > 6)?

c. What is the probability that he gets five, six, or seven hits.

d. What is the probability that he gets exactly six hits (hits = 6)?

*** Show your work please ***

Explanation / Answer

Solution:-

p = 0.30

n = 20

a) The expected number of hits for the week is 6.

E(x) = n × p

E(x) = 20 × 0.30

E(x) = 6

b) The probability that he gets more than six hits (hits > 6) is 0.392.

x = 6, p = 0.30, n = 20

By applying binomial distribution

P(x,n) = nCx*px*(1-p)(n-x)

P(x > 6) = 0.392

c) The probability that he gets five, six, or seven hits is 0.535.

x = 5, 6, 7 p = 0.30, n = 20

By applying binomial distribution

P(x,n) = nCx*px*(1-p)(n-x)

P(x = 5, 6, 7) = P(x = 5) + P(x = 6) + P(x = 7)

P(x = 5, 6, 7) = 0.179 + 0.192 + 0.164

P(x = 5, 6, 7) = 0.535

d) The probability that he gets exactly six hits is 0.192

x = 6, p = 0.30, n = 20

By applying binomial distribution

P(x,n) = nCx*px*(1-p)(n-x)

P(x = 6) = 0.192

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