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Conditional Probability

Name: Anonymous 2009-10-30 12:51

Ok /sci/, inb4 do your own homework, I really don't know why I can't get the solution to this one, even though I know it's dead simple, I've just been going around in circles.

A factory manufactures computer chips. 5% are defective. A Machine tests each chip. with probability 7*10^-3 that it incorrectly passes a defective chip, and probability 3*10^-4 that it incorrectly rejects a working chip. What is the probability that the machine will pass a random chip?

take P(D) to be the probability a chip is defective, and P(P)to be the probability the machine passes the chip. And for ease of writing, lets say P(_D) represents P(not D)
So we have
 P(D)=1/20
P(_D)=19/20
P(P intersect D)=7*10^-3
P(_P intersect _D)=3*10^-4
I've already worked out that P(D Given P)=0.14,
and P(_D Given _P)=3.157..*10^-4)

I really can't see a clear way of getting the solution.

P.S I'm a physics and Maths Joint Honours student, I have to do this Statistics module, but I've always hated Statistics.
Halp!

Name: 4tran 2009-10-30 23:49

if it passes a chip,
correctly passes a good chip or
incorrectly passes a bad chip

Name: Anonymous 2009-10-31 4:26

incorrectly pass: 0.05*7E-3
correctly pass: 0.95*(1-7E-3)
0.05*7E-3 + 0.95*(1-7E-3) = 0.9437

Name: Anonymous 2009-10-31 10:14

Cheers.

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