From 27 pieces of luggage, an airline luggage handler damages a random sample of four. the probability that exactly one of the damaged pieces of luggage is insured is twice the probability that none of the damaged pieces are insured. calculate the probability that exactly two of the four damaged pieces are insured.

0.273

Step-by-step explanation:

Let the number of insured pieces of luggage be i and u be the number of uninsured pieces of luggage, therefore,

i + u = 27

Now,

probability that exactly one of the damaged pieces of luggage is insured = (iC1)(uC3)/(27C4)

probability that none of the damaged pieces are insured = (uC4)/(27C4)

and,

(iC1)(uC3)/(27C4) = 2 (uC4)/(27C4)

=> u − 2i = 3

By solving, i + u = 27 and  u − 2i = 3

i = 8 and u = 19

and,

(8C2)(19C2)/(27C4) = 0.273

i think the answer is .36

step-by-step explanation:

what i did was only 2.88 divided by 8

135 is your answer

step-by-step explanation:

try 180 times three then divide that number by 4 (sides) to get your answer. im glad to : )