SOLUTION: Please help me solve the following problem as soon as possible. Thank you very much. A plant has 4 workshops. The production of workshop 1,2,3 & 4 account for 1/3, 1/4, 1/4 and

Question 252282: Please help me solve the following problem as soon as possible. Thank you very much.
A plant has 4 workshops. The production of workshop 1,2,3 & 4 account for 1/3, 1/4, 1/4 and 1/6 ,respectively, of the total output of the plant. The error rates of defective products correspond to the 4 workshops are 0.15, 0.08, 0.05, 0.01 respectively.
a)Find the probability to extract one random product that is a good product.
b)Find the probability to extract 7 random products that have 2 defective products.
Answer by edjones(8007) (Show Source): You can put this solution on YOUR website!
a)
1/3 * .15 = .05
1/4 * .08 = .02
1/4 * .05 = .0125
1/6 * .01 = .0167
-------------------------
=========>.08417
.
1-.08417=.91583 prob 1 random product is a good one.
.
.
b)
Let g=.91583 and d=.08417
(g+d)^7
nCr = n!/((n-r)r!) to find the coefficient of (g^5d^2)
7C2=21
21*.91583^5*.08417^2=.09585 prob that 2 out of 7 randomly selected will be defective.
.
Ed
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