# math

A factory finds out that on the average 20% of bolts produced by the given machine will be defective.if 10 bolts are selected at random from the day's production of this machine,find the probability that 1.exactly 2 will be defective.

2. 2 or more will be defective
3. More than 5 will be defective

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1. Prob(defect) = .2
prob(not defective = .8

prob(exactly 2 of 10 are defective_
= C(10,2) (.2)^2 (.8)^8
= ....

2. Use the "back-door" approach

if 2 or more are defective, then
we can't have:
none defective or only 1 defective.
= 1 - C(10,0) (.2)^0 (.8)^10 - C(10,1) (.2)(.8^9
= ...

3. For this you will just have to grind out
C(10,5) (.2)^5 (.8)^5 + C(10,6) (.2)^6 (.8)^4 + ... + C(10,10) .2^10 .8^0

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posted by Reiny

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