Question

Kenny has a 1/3 probability of getting a hit and a 1/6 probability of drawing a base-on-balls every time he appears at the plate. Suppose he negotiates his contract for an end-of-the-season bonus of $1,000 for each hit and $100 for each base-on-balls he gets. If he can reasonably expect 600 plate appearances this year, how much does he expect his bonus will be?

Answer 1

Answer:

He expects that his bonus will be $210,000.

Step-by-step explanation:

For each plate appearence, there are only two possible outcomes. Either he gets on base(base hit or bases-on-balls), or he does not. The probability of getting on base on each plate appearence is independent of other plate appearences. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

The expected value of the binomial distribution is:

$E(X) = np$

600 plate appearances this year

So n = 600.

Expected number of hits

$p = \frac{1}{3}$

So

$E(X) = np = 600\frac{1}{3} = 200$

Expected number of base on balls.

$p = \frac{1}{6}$

So

$E(X) = np = 600\frac{1}{6} = 100$

Bonus

$1,000 for each hit and $100 for each base-on-balls he gets.

200*1,000 + 100*100 = 210,000

He expects that his bonus will be $210,000.