A firm places three orders for supplies among five different distributors. Each order is randomly assigned to one of the distributors, and a distributor may receive multiple orders.

Find the probabilities of the following events.

a. All orders go to different distributors.

b. All orders go to the same distributor.

c. Exactly two of the three orders go to one particular distributor.

a) 12/25

b) 1/25

c) 12/25

Step-by-step explanation:

a) The first order can go anywhere, so it has 5 possibilities out of 5. The second order can go to any distributor except the one that got the first order, so it has 4 possibilities out of 5. The third order may go to any of the 3 remaining distributors, giving us 3 chances out of 5. So the probability is 5/5*4/5*3/5 = 12/25

b) The first order, again, can go anywhere, so it has 5 out of 5. The second and third order have to go to the same distributor, so, in each case, we have only 1 option available out of 5. The probability of this event is 5/5*1/5*1/5 = 1/25

c) This event is the complementary event of the union of the other 2 events the exercise has. If neither the same distributor got all 3 orders non all orders go to different distributors, then necessarily one distributor should got 2 orders and other 1. The probability of this event is, therefore 1-12/25-1/25 = 12/25.