Respuesta :
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:
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:
[tex] \\ P(X) = 24\% = 0.24[/tex]
The probability that the assembling company receives a part from supplier Y:
[tex] \\ P(Y) = 36\% = 0.36[/tex]
The probability that the assembling company receives a part from supplier Z:
[tex] \\ P(Z) = 40\% = 0.40[/tex]
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:
[tex] \\ P(D|X) = 5\% = 0.05[/tex]
And the probabilities that defective parts come from Y and Z are respectively:
[tex] \\ P(D|Y) = 10\% = 0.10[/tex]
[tex] \\ P(D|Z) = 6\% = 0.06[/tex]
So, having all these probabilities at hand, we can solve the question using the formula for Bayes' Theorem:
[tex] \\ P(Z|D) = \frac{P(Z)*P(D|Z)}{P(X)*P(D|X) + P(Y)*P(D|Y) + P(Z)*P(D|Z)}[/tex]
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:
[tex] \\ P(Z|D) = \frac{0.40*0.06}{0.24*0.05 + 0.36*0.10 + 0.40*0.06}[/tex]
[tex] \\ P(Z|D) = \frac{0.024}{0.072}[/tex]
[tex] \\ P(Z|D) = \frac{1}{3} = 0.3333... = 33.33\%[/tex]
So, the probability that a computer has a defective part in it and comes from supplier Z is about 33.33%.
