Question

A company orders supplies from M distributors and wishes to place n orders (n < M). Assume that the company places the orders in a manner that allows every distributor an equal chance of obtaining any one order and there is no restriction on the number of orders that can be placed with any distributor. Suppose that the number of distributors is M = 10 and that there are n = 8 orders to be placed. (Round your answers to four decimal places.)

(a) What is the probability that all of the orders go to different distributors?

(b) What is the probability that distributor I gets exactly two orders and distributor II gets exactly three orders?

(c) What is the probability that distributors I, II, and III get exactly two, three, and one order(s), respectively?

Answer 1

a)

here for each order there are 10  distributors ; number of ways so that 8 orders to be placed

=10⁸ =100000000

number of ways so that 8 orders can be placed to different distributors =N(select 8 out of 10 and then arrange them in order )=¹⁰C₈*8! =45*40320=1814400

hence P(all of the orders go to different distributors) =1814400/100000000=0.018144

b)

number of ways  that distributor I gets exactly two orders and distributor II gets exactly three orders

⁸C₂*⁶C₃*8³ =286720

hence probabiltiy =286720/100000000=0.002867

c)

number of ways distributors I, II, and III get exactly two, three, and one order=⁸C₂*⁶C₃*³C₁*7²

=41160

hence probabiltiy =41160/100000000=0.0004116