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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 = 7 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 :

aliakbar921853

Answer:

a) 0.06048 b) 0.001344 c) 0.000294

Step-by-step explanation:

number of distributors = 10;

number of orders = 7

a)

There are 10 distributors for each order, so the no. of ways the 7 orders could be placed are = 10⁷ =10000000

no. of ways that orders goes to different distributors = ¹⁰C₇*7! = 604800

Probability = 604800/10000000 = 0.06048

b)

number of ways distributor 1 gets exactly 2 and distributor 2 gets exactly 3 orders = ⁷C₂*⁵C₃*8² = 13440

probability = 13440/10000000 = 0.001344

c)

number of ways distributor 1, 2,3 get exactly 2,3 and 1 order(s) = ⁷C₂*⁵C₃*²C₁*7¹ = 2940

probability = 2940/10000000 = 0.000294

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