Question

When Mr. Billy taught high school he once volunteered to allow students to throw pie tins of shaving cream at his face (with his head sticking through a hole in a board). Whether or not a pie tin hit him was independent of who was throwing and how many times they had thrown, with a probability of 0.08 that he would be hit. What is the probability that he was first hit on the 12th attempt?

Answer 1

3.20% probability that he was first hit on the 12th attempt

Explanation:

For each throw, there are only two possible outcomes. Either there is a hit, or there is not. Each throw is independent of each other. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

In which
`C_(n,x)` is the number of different combinations of x objects from a set of n elements, given by the following formula.

`C_(n,x) = (n!)/(x!(n-x)!)`

And p is the probability of X happening.

Probability of 0.08 that he would be hit.

This means that
`p = 0.08`

What is the probability that he was first hit on the 12th attempt?

No hits during the first 11 attempts(P(X = 0) when n = 11).

Hit during the 12th attempt, with 0.08 probability. So

`p = 0.08*P(X = 0)`

In which

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 0) = C_(11,0).(0.08)^(0).(0.92)^(11) = 0.3996`

`p = 0.08*P(X = 0) = 0.08*0.3996 = 0.0320`

3.20% probability that he was first hit on the 12th attempt