I believe this is a hypergeometric problem since we are assuming without replace

Question

I believe this is a hypergeometric problem since we are assuming without replacement. I need help with how to actually write and solve this problem. Thanks.

Problem 1 A batch of parts contains 100 parts from Supplier A and 200 parts from Supplier B a. b. c. If four parts are selected what is the probability they are all from Supplier A? If six parts are selected what is the probability that at least three are from Supplier A? If ten parts are selected what is the probability that at least one is from Supplier B?

Explanation / Answer

P(A) = 1/3

P(B) = 2/3

a. P(All 4 from supplier A) = (1/3)^4 = 0.0039

b. P(Atleast 3 from 6 from A) = C(6,3)*(1/3)^3*(2/3)^3 +C(6,4)*(1/3)^4*(2/3)^2 +C(6,5)*(1/3)^5*(2/3)^1 +C(6,6)*(1/3)^6

=0.313

c.

P(Atlease 1 from supplier B out of 10) = 1- P(all from supplier A) = 1-(1/3)^10 = 0.99998

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