Respuesta :
Answer:
The probability of one or more catastrophes in:
(1) Two mission is 0.0166.
(2) Five mission is 0.0410.
(3) Ten mission is 0.0803.
(4) Fifty mission is 0.3419.
Step-by-step explanation:
Let X = number of catastrophes in the missions.
The probability of a catastrophe in a mission is, P (X) = [tex]p=\frac{1}{120}[/tex].
The random variable X follows a Binomial distribution with parameters n and p.
The probability mass function of X is:
[tex]P(X=x)={n\choose x}\frac{1}{120}^{x}(1-\frac{1}{120})^{n-x};\x=0,1,2,3...[/tex]
In this case we need to compute the probability of 1 or more than 1 catastrophes in n missions.
Then the value of P (X ≥ 1) is:
P (X ≥ 1) = 1 - P (X < 1)
= 1 - P (X = 0)
[tex]=1-{n\choose 0}\frac{1}{120}^{0}(1-\frac{1}{120})^{n-0}\\=1-(1\times1\times(1-\frac{1}{120})^{n-0})\\=1-(1-\frac{1}{120})^{n-0}[/tex]
(1)
Compute the compute the probability of 1 or more than 1 catastrophes in 2 missions as follows:
[tex]P(X\geq 1)=1-(1-\frac{1}{120})^{2-0}=1-0.9834=0.0166[/tex]
(2)
Compute the compute the probability of 1 or more than 1 catastrophes in 5 missions as follows:
[tex]P(X\geq 1)=1-(1-\frac{1}{120})^{5-0}=1-0.9590=0.0410[/tex]
(3)
Compute the compute the probability of 1 or more than 1 catastrophes in 10 missions as follows:
[tex]P(X\geq 1)=1-(1-\frac{1}{120})^{10-0}=1-0.9197=0.0803[/tex]
(4)
Compute the compute the probability of 1 or more than 1 catastrophes in 50 missions as follows:
[tex]P(X\geq 1)=1-(1-\frac{1}{120})^{50-0}=1-0.6581=0.3419[/tex]
