Respuesta :
Answer:
[tex]P=0.1833[/tex]
Step-by-step explanation:
The number of ways or combinations in which we can choose x elements from a group of n is calculated as:
[tex]nCx=\frac{n!}{x!(n-x)!}[/tex]
So, the probability that none of the damaged houses are insured can be calculated as:
[tex]\frac{kC3}{10C3}=\frac{kC3}{120}[/tex]
Because there are 120 ways to select 3 houses from the block of 10 houses and there are kC3 ways to select 3 houses from a group of k houses that are not insured.
The probability that none of the damaged houses are insured is 1/120, so:
[tex]\frac{kC3}{120}=\frac{1}{120}\\ kC3=1[/tex]
kC3 is equal to 1, only if k is equal to 3. It means that there are 3 houses that are not insured.
Then, the probability that one of the damaged houses is insured can be calculated as:
[tex]\frac{3C2*7C1}{10C3}=0.175[/tex]
Because we select 2 houses from the 3 that are not insured and select 1 house from the 7 that are insured.
Finally, the probability that at most one of the damaged houses is insured is equal to the sum of the probability that one of the damaged houses is insured and the probability that none of the damaged houses is insured. It is equal to:
[tex]P=\frac{1}{120}+0.175=0.1833[/tex]
