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Probability question regarding independence

Usagi

Member
Jun 26, 2012
46
A supplier sends boxes of screws to a factory: 90% of the boxes contain 1% defective, 9% contain 10% defective, and 1% contain 100% defective (eg wrong size).

i) What percentage of screws supplied are defective?

ii) Two screws are chosen from a randomly selected box. What is the probability that both are defective?

iii) Given that both are defective, what is the probability that the box is 100% defective?

I have attempted all 3 parts but I am unsure if my working is correct, would someone be kind enough to verify my working/reasoning? Thank you.

i) Let D be the event that the screws supplied are defective. Let $A_1$ be the event that the box containing 1% defective was sent to the factory. Likewise, let $A_2$ and $A_3$ be the event that the box containing 10% and 100% defective, respectively, that was sent to the factory.

Thus using the total probability theorem we have:

$P(D) = P(A_1)P(D|A_1) + P(A_2)P(D|A_2)+P(A_3)P(D|A_3) = (0.9)(0.01)+0.09(0.1)+0.01(1) = 0.028$

ii) Let Z be the event that two screws are both defective and let $D_i$ be the event that the ith screw is defective from a random box. Thus using the total probability theorem again we have:

$P(Z) = P(A_1)P(Z|A_1)+P(A_2)P(Z|A_2) + P(A_3)P(Z|A_3) = P(A_1)P(D_1 \cap D_2|A_1) + P(A_2)P(D_1 \cap D_2|A_2) + P(A_3)P(D_1 \cap D_2|A_3)$

$= P(A_1)P(D_1|A_1)P(D_2|A_1) + P(A_2)P(D_1|A_2)P(D_2|A_2) + P(A_3)P(D_1|A_3)P(D_2|A_3)$ because we can assume $D_1$ and $D_2$ are conditionally independent of $A_i$'s.

$=0.01(0.01)^2+0.09(0.1)^2+0.01(1)^2 = 0.010901$

iii) We need to find $P(A_3|Z) = \frac{P(A_3)P(Z|A_3)}{P(Z)} = \frac{0.01(1)}{0.010901}$

Any help is greatly appreciated!
 

CaptainBlack

Well-known member
Jan 26, 2012
890
A supplier sends boxes of screws to a factory: 90% of the boxes contain 1% defective, 9% contain 10% defective, and 1% contain 100% defective (eg wrong size).

i) What percentage of screws supplied are defective?

ii) Two screws are chosen from a randomly selected box. What is the probability that both are defective?

iii) Given that both are defective, what is the probability that the box is 100% defective?

I have attempted all 3 parts but I am unsure if my working is correct, would someone be kind enough to verify my working/reasoning? Thank you.

i) Let D be the event that the screws supplied are defective. Let $A_1$ be the event that the box containing 1% defective was sent to the factory. Likewise, let $A_2$ and $A_3$ be the event that the box containing 10% and 100% defective, respectively, that was sent to the factory.

Thus using the total probability theorem we have:

$P(D) = P(A_1)P(D|A_1) + P(A_2)P(D|A_2)+P(A_3)P(D|A_3) = (0.9)(0.01)+0.09(0.1)+0.01(1) = 0.028$

ii) Let Z be the event that two screws are both defective and let $D_i$ be the event that the ith screw is defective from a random box. Thus using the total probability theorem again we have:

$P(Z) = P(A_1)P(Z|A_1)+P(A_2)P(Z|A_2) + P(A_3)P(Z|A_3) = P(A_1)P(D_1 \cap D_2|A_1) + P(A_2)P(D_1 \cap D_2|A_2) + P(A_3)P(D_1 \cap D_2|A_3)$

$= P(A_1)P(D_1|A_1)P(D_2|A_1) + P(A_2)P(D_1|A_2)P(D_2|A_2) + P(A_3)P(D_1|A_3)P(D_2|A_3)$ because we can assume $D_1$ and $D_2$ are conditionally independent of $A_i$'s.

$=0.01(0.01)^2+0.09(0.1)^2+0.01(1)^2 = 0.010901$

iii) We need to find $P(A_3|Z) = \frac{P(A_3)P(Z|A_3)}{P(Z)} = \frac{0.01(1)}{0.010901}$

Any help is greatly appreciated!
For part ii you should have: \(0.9(0.01)^2 + 0.09(0.1)^2+0.01(1)^2\)


CB
 

Usagi

Member
Jun 26, 2012
46
Oh yes, silly me, typo'ed.

Thanks for the confirmation :)