Respuesta :
Answer:
[tex]P(x \geq 3)=1- [P(X=0)+P(X=1) +P(X=2)][/tex]
And we can find the individual probabilities like this:
[tex]P(X=0)=(15C0)(0.08)^0 (1-0.08)^{15-0}=0.286[/tex]
[tex]P(X=1)=(15C1)(0.08)^1 (1-0.08)^{15-1}=0.373[/tex]
[tex]P(X=2)=(15C2)(0.08)^1 (1-0.08)^{15-2}=0.227[/tex]
[tex]P(x \geq 3)=1- [0.286+0.373+0.227]=0.114[/tex]
Step-by-step explanation:
Let X the random variable of interest "number of cars involved in an accident", on this case we now that:
[tex]X \sim Binom(n=15, p=0.08)[/tex]
The probability mass function for the Binomial distribution is given as:
[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]
Where (nCx) means combinatory and it's given by this formula:
[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]
And we want to find this probability:
[tex]P(x \geq 3)=1- [P(X=0)+P(X=1) +P(X=2)][/tex]
And we can find the individual probabilities like this:
[tex]P(X=0)=(15C0)(0.08)^0 (1-0.08)^{15-0}=0.286[/tex]
[tex]P(X=1)=(15C1)(0.08)^1 (1-0.08)^{15-1}=0.373[/tex]
[tex]P(X=2)=(15C2)(0.08)^1 (1-0.08)^{15-2}=0.227[/tex]
[tex]P(x \geq 3)=1- [0.286+0.373+0.227]=0.114[/tex]
