prob(fail) = .2
prob(not fail) = .8
prob(exactly 2 of 10 will fail)
= C(10,2) .2^2 .8^8
= appr .302
2 or more defective ---> exclude 0 not fail and 1 not fail
= 1 - C(10,0) .8^0 .2^10 - C(10,1) (.8)( .2^9)
= ....
more than 5
= prob(exactly 6) + prob(exactly 7) + ... + prob(exactly 10)
= ...
just grind that one out
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
(base on probability and combinatorial analysis)
1 answer
1 answer