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Word problem using Bayes' theorem.

Linty Fresh

Junior Member
Joined
Sep 6, 2005
Messages
58
At an electronics plant, it's known from past experience that the probability is 0.86 that a new worker who has attended the company's training program will meet his production quota, and that the corresponding probability is 0.35 for a new worker who has not attende the company's training program. If 80% of all new workers attend the training program, what is the probability that

(a) a new worker will not meet his production quota
(b)a new worker who does not meet his production quota will not have attended the company's training program?

I got (a) easily enough:

P(Q|T)=0.86
p(Q|T')=0.35
P(T)=0.80

So the first branch on our way to get P(Q) is P(T) x P(Q|T) = 0.8 X 0.86 = 0.688

and the second branch is P(T') X P(Q|T') = 0.2 x P(Q|T') = 0.2 x 0.35 = 0.07

To get P(Q), we add 0.688 and 0.07 to get 0.758, so P(Q') = 1 - 0.758 = 0.242.

But how do I get (b)? How does one get P(T'|Q') from any of this? OK, what if I did this:

P(Q'|T')= P(Q') x P(T'|Q')/P(T') . . . but how do I know P(T'|Q')? I know I'm missing something that's right in front of me. Is it something to do with P(T'?Q')?
 
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