Question

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?

Answer 1

Answer:

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

Step-by-step explanation:

This can be formulated as the following problem:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula

$P = \frac{P(B).P(A/B)}{P(A)}$

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:

$P(A) = P_{1} + P_{2} + P_{3}$

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

$P_{1} = 0.24*0.05 = 0.012$

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

$P_{2} = 0.36*0.10 = 0.036$

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

$P_{2} = 0.40*0.06 = 0.024$

So

$P(A) = P_{1} + P_{2} + P_{3} = 0.012 + 0.036 + 0.024 = 0.072$

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:

$P = \frac{0.4*0.06}{0.072} = 0.33$

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