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.

Answer:0.273Step-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 = 3By solving, i + u = 27 and  u − 2i = 3i = 8 and u = 19and,(8C2)(19C2)/(27C4) = 0.273