Question

The ski lift at Thunder Mountain called Devils Canyon is quite dangerous. Usually 2% of the skiers whoattempt the run will injure themselves due to falls. What is the probability that no more than 5 of the next 100 skiers who attempt this run will injure themselves. Answer Choices. A. 0.1867. B. 0.7652. C. 0.8899. D. 0.9845

Answer 1

Given that, there is a 2% of probability that a skier that attempts to run will injure himself on a fall.

You have to calculate the probability that no more than 5 skiers hurt themselves due to a fall in the next 100 skiers that will attempt the run.

This scenario describes a binomial experiment, where:

- The number of trials is fixed: n=100

- The trials are independent of each other.

- The probability of success is constant throughout the experiment, p=0.02

You have to calculate the probability that 5 or fewer skiers get injured, you can express this probability as follows:

`P(X\leq5)`

The possible outcomes that are included in this expression are X=5, X=4, X=3, X=2, X=1, and X=0.

So, the probability can be expressed as follows:

`P(X\leq5)=P(X=5)+P(X=4)+P(X=3)+P(X=2)+P(X=1)+P(X=0)`

Using the formula of the binomial probability, you can determine each one of the probabilities listed above.

`P(X=x)=(n!)/((n-x)!x!)*p^x*q^((n-x))`

Where

p is the probability of success

q is the probability of failure

n is the sample size

x is the number of successes

For this exercise:

p=0.02

q=1-p=1-0.02=0.98

n=100

-Calculate the probability of x=5

`\begin{gathered} P(X=5)=(100!)/((100-5)!5!)*0.02^5*0.98^((100-5)) \\ P(X=5)=(100!)/(95!*5!)*0.02^5*0.98^(95) \\ P(X=5)=0.0353 \end{gathered}`

-Calculate the probability of x=4

`\begin{gathered} P(X=4)=(100!)/((100-4)!4!)*0.02^4*0.98^((100-4)) \\ P(X=4)=(100!)/(96!*4!)*0.02^4*0.98^(96) \\ P(X=4)=0.0902 \end{gathered}`

-Calculate the probability of x=3

`\begin{gathered} P(X=3)=(100!)/((100-3)!)*0.02^3*0.98^((100-3)) \\ P(X=3)=(100!)/(97!*3!)*0.02^3*0.98^(97) \\ P(X=3)=0.1823 \end{gathered}`

-Calculate the probability of x=2

`\begin{gathered} P(X=2)=(100!)/((100-2)!2!)*0.02^2*0.98^((100-2)) \\ P(X=2)=(100!)/(98!*2!)*0.02^2*0.98^(98) \\ P(X=2)=0.2734 \end{gathered}`

-Calculate the probability of x=1

`\begin{gathered} P(X=1)=(100!)/((100-1)!1!)*0.02^1*0.98^((100-1)) \\ P(X=1)=(100!)/(99!*1!)*0.02^1*0.98^(99) \\ P(X=1)=0.2707 \end{gathered}`

-Calculate the probability of x=0

`\begin{gathered} P(X=0)=(100!)/((100-0)!*0!)*0.02^0*0.98^((100-0)) \\ P(X=0)=(100!)/(100!*0!)*0.02^0*0.98^(100) \\ P(X=0)=0.1326 \end{gathered}`

Now you can determine the probability of 5 or fewer skiers getting injured in a fall:

`\begin{gathered} P(X\leq5)=P(X=5)+P(X=4)+P(X=3)+P(X=2)+P(X=1)+P(X=0) \\ P(X\leq5)=0.0353+0.0902+0.1823+0.2734+0.2707+0.1326 \\ P(X\leq5)=0.9845 \end{gathered}`

The probability that no more than 5 skiers will suffer an injury is 0.9845.