Respuesta :
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:
[tex]P(X\leq5)[/tex]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:
[tex]P(X\leq5)=P(X=5)+P(X=4)+P(X=3)+P(X=2)+P(X=1)+P(X=0)[/tex]Using the formula of the binomial probability, you can determine each one of the probabilities listed above.
[tex]P(X=x)=\frac{n!}{(n-x)!x!}*p^x*q^{(n-x)}[/tex]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
[tex]\begin{gathered} P(X=5)=\frac{100!}{(100-5)!5!}*0.02^5*0.98^{(100-5)} \\ P(X=5)=\frac{100!}{95!*5!}*0.02^5*0.98^{95} \\ P(X=5)=0.0353 \end{gathered}[/tex]-Calculate the probability of x=4
[tex]\begin{gathered} P(X=4)=\frac{100!}{(100-4)!4!}*0.02^4*0.98^{(100-4)} \\ P(X=4)=\frac{100!}{96!*4!}*0.02^4*0.98^{96} \\ P(X=4)=0.0902 \end{gathered}[/tex]-Calculate the probability of x=3
[tex]\begin{gathered} P(X=3)=\frac{100!}{(100-3)!}*0.02^3*0.98^{(100-3)} \\ P(X=3)=\frac{100!}{97!*3!}*0.02^3*0.98^{97} \\ P(X=3)=0.1823 \end{gathered}[/tex]-Calculate the probability of x=2
[tex]\begin{gathered} P(X=2)=\frac{100!}{(100-2)!2!}*0.02^2*0.98^{(100-2)} \\ P(X=2)=\frac{100!}{98!*2!}*0.02^2*0.98^{98} \\ P(X=2)=0.2734 \end{gathered}[/tex]-Calculate the probability of x=1
[tex]\begin{gathered} P(X=1)=\frac{100!}{(100-1)!1!}*0.02^1*0.98^{(100-1)} \\ P(X=1)=\frac{100!}{99!*1!}*0.02^1*0.98^{99} \\ P(X=1)=0.2707 \end{gathered}[/tex]-Calculate the probability of x=0
[tex]\begin{gathered} P(X=0)=\frac{100!}{(100-0)!*0!}*0.02^0*0.98^{(100-0)} \\ P(X=0)=\frac{100!}{100!*0!}*0.02^0*0.98^{100} \\ P(X=0)=0.1326 \end{gathered}[/tex]Now you can determine the probability of 5 or fewer skiers getting injured in a fall:
[tex]\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}[/tex]The probability that no more than 5 skiers will suffer an injury is 0.9845.
