A baseball player has a batting average of 0.260. Suppose that you observe succe

Question

A baseball player has a batting average of 0.260. Suppose that you observe successive at-bats of the player and note for each at-bat, whether the player gets a hit. You may assume that each at-bat is independent of one another.

1) Find the probability that the SECOND hit by the player occurs on his FIFTH at-bat.

2) Find the probability that the SECOND hit by the player occurs BETWEEN his THIRD and SIXTH at-bats, inclusive.

3) Find the MEAN number of at-bats needed to get the SECOND hit.

Thanks for your help!

Explanation / Answer

1) There are two possible outcomes:hit or no hit. Therefore, the distribution follows geometric one, with probability of success, p=0.260 and 1-p=0.74, the probbaility of failure.

P( X=2)=(0.260)^1(0.74)^1=0.1924 [P(X=r)=p^(r-1)*(1-p)]

2) P(X=3)=(0.260)^2(0.74)^1=0.05 and P(X=6)=(0.260)^5(0.74)^1=0.0009

Ans: 0.05*0.0009=0.000045

3) E(X)=1/p=1/0.1924=5.20 [E(X)=1/p]

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