Question

1. A luge race is very dangerous, and a crash can cause serious injuries. The league requires anyone who has a crash to have a thorough medical screening before they are allowed to race again. A certain performer has an independent .04 probability of a crash in each race. a) What is the probability she will have her first crash within the first 30 races she runs this season

Answer 1

0.7061 = 70.61% probability she will have her first crash within the first 30 races she runs this season

Explanation:

For each race, there are only two possible outcomes. Either the person has a crash, or the person does not. The probability of having a crash during a race is independent of whether there was a crash in any other race. This means that the binomial probability distribution is used 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.

A certain performer has an independent .04 probability of a crash in each race.

This means that
`p = 0.04`

a) What is the probability she will have her first crash within the first 30 races she runs this season

This is:

`P(X \geq 1) = 1 - P(X = 0)`

When
`n = 30`

We have that:

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

`P(X = 0) = C_(30,0).(0.04)^(0).(0.96)^(30) = 0.2939`

`P(X \geq 1) = 1 - P(X = 0) = 1 - 0.2939 = 0.7061`

0.7061 = 70.61% probability she will have her first crash within the first 30 races she runs this season