A computer assembling company receives 24% of parts from supplier X, 36% of parts from supplier Y, and the remaining 40% of parts from supplier Z. Five percent of parts supplied by X, ten percent of parts supplied by Y, and six percent of parts supplied by Z are defective. If an assembled computer has a defective part in it, what is the probability that this part was received from supplier Z?

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

-In your problem, we have:

P(A) is the probability of a defective part being supplied. For this probability, we have:

[tex]P(A) = P_{1} + P_{2} + P_{3}[/tex]

In which [tex]P_{1}[/tex] is the probability that the defective product was chosen from supplier X(we have to consider the probability of supplier X being chosen). So:

[tex]P_{1} = 0.24*0.05 = 0.012[/tex]

[tex]P_{2}[/tex] is the probability that the defective product was chosen from supplier Y(we have to consider the probability of supplier Y being chosen). So:

[tex]P_{2} = 0.36*0.10 = 0.036[/tex]

[tex]P_{3}[/tex] is the probability that the defective product was chosen from supplier Z(we have to consider the probability of supplier Z being chosen). So:

[tex]P_{2} = 0.40*0.06 = 0.024[/tex]

So

[tex]P(A) = P_{1} + P_{2} + P_{3} = 0.012 + 0.036 + 0.024 = 0.072[/tex]

P(B) is the probability of the supplier chosen being Z, so P(B) = 0.4

P(A/B) is the probability of the part supplied being defective, knowing that the supplier chosen was Z. So P(A/B) = 0.06.

So, the probability that this part was received from supplier Z is:

[tex]P = \frac{0.4*0.06}{0.072} = 0.33[/tex]

There is a 33% probability that this party was received from supplier Z.