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

Answer:

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

Step-by-step 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.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

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