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Last season Ryan hit a homerun in about $7\%$ of his bats. Suppose we model at bat as the outcome of Bernoulli random variable. In a typical week, Ryan takes $25$ at bats.

consider the following gamble. In a given week you have to pay your friend $\$5$ for each at bat and he pays you $\$25$ for each home run, plus he pays you an additional bonus of $\$80$, if Ryan hits more than $3$ home runs. Would you expect to make money on such a game?

I thought that the cost for me is $5\cdot 25 =125$ and the earning for me I get $25\cdot\frac{7}{4}$ plus something I don’t get it from here. I mean I don't know where to put $\$80$.

Please help.

MattAllegro
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lee
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1 Answers1

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Your expected costs are $E(c)=25\cdot 0.07\cdot 5$ (expected value).

$25\cdot 0.07$ are the expected homeruns. It is $n\cdot p$. This is the expected value of the binomial distribution.

Your expected earnings are $E(e)=P(X > 3)\cdot 80$

X is the binomial distributed random variable with $p=0.07$ and $n=25$.

It is helpful to know for calculation, that

$P(X > 3)=1-P(X=0)-P(X=1)-P(X=2)-P(X=3)$

callculus42
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