Question

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

Answer 1

Answer:

thus the probability that a part was received from supplier Z , given that is defective is 5/6 (83.33%)

Step-by-step explanation:

denoting A= a piece is defective , Bi = a piece is defective from the i-th supplier and Ci= choosing a piece from the the i-th supplier

then

P(A)= ∑ P(Bi)*P(C) with i from 1 to 3

P(A)= ∑ 5/100 * 24/100 + 10/100 * 36/100 + 6/100 * 40/100 = 9/125

from the theorem of Bayes

P(Cz/A)= P(Cz∩A)/P(A)

where

P(Cz/A) = probability of choosing a piece from Z , given that a defective part was obtained

P(Cz∩A)= probability of choosing a piece from Z that is defective = P(Bz) = 6/100

therefore

P(Cz/A)= P(Cz∩A)/P(A) = P(Bz)/P(A)= 6/100/(9/125) = 5/6 (83.33%)

thus the probability that a part was received from supplier Z , given that is defective is 5/6 (83.33%)

Answer 2

Answer:

The probability that a computer has a defective part in it and comes from supplier Z is about 33.33%.

Step-by-step explanation:

This question can be solved using the Bayes' Theorem and with it, we can calculate conditional probabilities, that is, the probability that an event A can occur given that an event B has occurred previously (roughly speaking).

To solve the question, we need to identify each of the probabilities given:

The probability that the assembling company receives a part from supplier X is:

$\\ P(X) = 24\% = 0.24$

The probability that the assembling company receives a part from supplier Y:

$\\ P(Y) = 36\% = 0.36$

The probability that the assembling company receives a part from supplier Z:

$\\ P(Z) = 40\% = 0.40$

We also have the probabilities that a defective part is supplied, respectively, by X, Y, and Z, so we can write them as conditional probabilities:

The probability that a defective part comes from X is:

$\\ P(D|X) = 5\% = 0.05$

And the probabilities that defective parts come from Y and Z are respectively:

$\\ P(D|Y) = 10\% = 0.10$

$\\ P(D|Z) = 6\% = 0.06$

So, having all these probabilities at hand, we can solve the question using the formula for Bayes' Theorem:

$\\ P(Z|D) = \frac{P(Z)*P(D|Z)}{P(X)*P(D|X) + P(Y)*P(D|Y) + P(Z)*P(D|Z)}$

Notice that the denominator permits us to calculate the total probability for finding a defective part, and the numerator of this fraction, the portion of the probability that corresponds to the part that comes from supplier Z.

Thus, substituting each probability accordingly:

$\\ P(Z|D) = \frac{0.40*0.06}{0.24*0.05 + 0.36*0.10 + 0.40*0.06}$

$\\ P(Z|D) = \frac{0.024}{0.072}$

$\\ P(Z|D) = \frac{1}{3} = 0.3333... = 33.33\%$

So, the probability that a computer has a defective part in it and comes from supplier Z is about 33.33%.