Question

It's the bottom of the 9 inning of a ballgame, with 2 outs. The next three batters are Brandon, Francisco and Pete. Brandon reaches base 38% of the time. Francisco reaches base 34% of the time. Pete hits a homerun in 1/12 of the times he comes up. You may assume these events are independent of one another (though they likely wouldn't be). What are the chances that Brandon and Francisco both reach base, followed by Pete hitting a homerun to win the game? Give your answer in percents, rounded to the nearest tenth of a percent.

Answer 1

Given:

Probability that Brandon reaches the base = 0.38

Probability that Francisco reaches the base = 0.34

Probability that Pete hits a homerun = 1/12 or 0.0833

So, the chances that Brandon and Francisco both reach the base, followed by Pete hitting a home run to win the game is:

`P(\text{Brandon reaches base and Francisco the base) = P(Brandon reaches the base) }*\text{ }P(Francisco\text{ reaches the base)}`
`\begin{gathered} P(\text{Brandon reaches the base and Francisco reaches the base) = 038 }*\text{ 0.34 } \\ \text{ = 0.1292} \end{gathered}`

Hence the solution to the problem is 1.08%