Question

Five construction companies each offer bids on three DISTINCT Department of Transportation (DOT) contracts. A particular company will be awarded at most one DOT contract.

a) How many different ways can the bid be awarded?

b) Under the assumption that the simple events are equally likely, find the probability that company 2 is awarded a DOT contract.

c) Suppose that companies 4 and 5 have submitted non-competitive bids. If the contracts are awarded at random by the DOT, find the probability that both of these companies receive contracts.

Answer 1

(a) one company can have at most one contract

no. of ways 5*4*3 (in 1st contract 5 choices then in 2nd only 4 choices remain and then in 3rd contract only 3 choices remain)

no. of ways = 5*4*3 = 60

(b) choose 3 companies to get contract : company 2 already chosen then choose 2 companies out of remaining 4 (4C2)

ways to choose = (4C2) = 6

ways to arrange these 3 companies = 3! (since all three contracts distinct) = 6

total ways in which company 2 selected = 6*6 = 36

P(company 2 get contract) = 36/60 = 0.6

(c)

similar to (b)

choose 3 companies to get contract : company 4 and 5 already chosen then choose 1 companies out of remaining 3 (3C1)

ways to choose = (3C1) = 3

ways to arrange these 3 companies = 3! (since all three contracts distinct) = 6

total ways in which company 2 selected = 3*6 = 18

P(company 2 get contract) = 18/60 = 0.3

P.S. (please upvote if you find the answer satisfactory)