Acomputer 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?

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

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:

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

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

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

So

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:

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

it 2

step-by-step explanation:

8^1/3=2