3
$\begingroup$

Question 4.49 in Newbold (8. ed)

A company receives large shipments of parts from two sources. Seventy percent of the shipments come from a supplier whose shipments typically contain $10\%$ defectives, while the remainder are from a supplier whose shipments typically contain $20\%$ defectives. A manager receives a shipment but does not know the source. A random sample of $20$ items from this shipment is tested, and $1$ of the parts is found to be defective. What is the probability that this shipment came from the more reliable supplier? (Hint: Use Bayes' theorem.)

Using $P(A)=0.7$, $P(B)=0.3$, $P(D|A)=0.1$ and $P(D|B)=0.2$ I have made a two-way table and came to the conclusion that $P(A|D)=0.538$ and $P(B|D)=0.461$. So OK, a little over half of the defective parts come from supplier A (which makes sense, given they provide $70\%$ of the parts, even if their defect-rate is lower). However, I don't understand how all this relates to the actual question ("What is the probability that this shipment came from the more reliable supplier?"). From what could I calculate that? Could someone help, please?

$\endgroup$

1 Answer 1

1
$\begingroup$

The more reliable supplier is $A$ because its defection rate is smaller. And, we need the following probability: $$P(A|1D, 19D')=\frac{P(1D,19D'|A)P(A)}{P(1D,19D'|A)P(A)+P(1D,19D'|B)P(B)}$$

And, for example, one of the terms can be calculated as $$P(1D,19D'|A)={20 \choose 1}0.1^10.9^{19}$$

The rest can be calculated similarly and then substituted.

$\endgroup$
1
  • 1
    $\begingroup$ Thank you, @gunes. I've done it and now know that the probability that the shipment came from supplier A is 0.916225 :) (I always understand it (eventually), when you explain it but by myself I could never come up with these solutions... :| ) $\endgroup$
    – Reader 123
    Apr 6, 2020 at 18:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.