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Two companies, $A$ and $B$, produce the same objects.
- The objects that come from the company $A$ are defective with a probability of $0.05$;
- The objects that come from the company $B$ are defective with a probability of $0.01$;
Assuming that someone bought two objects from the same company, with a probability of $50 \%$. If the first object is defective, what is the probability that also the second one is defective?

I have calculated the probability to find the first defective object, that object can come from A or B:
In the formula:

  • $D_1$ denote the event "the first defective object";
  • $P(A)$ denote the probability to buy an object from the company $A$;
  • $P(B)$ denote the probability to buy an object from the company $B$;
  • $P(D_1|A)$ denote the probability to buy the first object knowing that come from $A$;
  • $P(D_1|B)$ denote the probability to buy the first object knowing that come from $A$;

$\begin{array}{rcl}P(D_1) & = & P(A) \cdot P(D_1 | A) + P(B) \cdot P(D_1|B) \\ & = & 50 \% \cdot 0.05 + 50 \% \cdot 0.01 \\ & = & \frac{0.05}{2} + \frac{0.01}{2} \\ & = & 0.03\end{array}$

but, I am unable to find the second part of the problem.

I have tried to do with a diagram, and I don't know if it is right, I think that those $0.05$ and $0.01$ are means, or approximations, and not exact quantities, i.e. in a bundle of 100 objects I can find 5 but also 0 or 1 etc... defective objects, or better if I consider a numerous bundle of 100 objects I can find a value that reaches 5 defective for each bundle. For this reason the second part of the diagram contains the same quantities, because I think that they are independent.

diagram-es10.png

Considering what I have done, the probability "also the second is defective given the first is defective" should be:

$\begin{array}{rcl}P(D_2|D_1) & = & 50 \% \cdot 0.05 \cdot 0.05 + 50 \% \cdot 0.01 \cdot 0.01 \\ & = & \frac{0.0025}{2} + \frac{0.0001}{2} \\ & = & 0.0013& \end{array}$

Please, can you help me? Thanks! :)

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  • $\begingroup$ @Peter Yes, but I have some difficulties to apply here. $\endgroup$
    – JB-Franco
    Sep 22, 2017 at 10:50
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    $\begingroup$ @Peter No need for Bayes' here. Just the law of total probability, as he has done on the last line. And to me, that seems entirely correct. $\endgroup$
    – Arthur
    Sep 22, 2017 at 10:55
  • $\begingroup$ @Arthur Thanks for the clarification $\endgroup$
    – Peter
    Sep 22, 2017 at 10:56
  • $\begingroup$ @Arthur It is not entirely correct. $\endgroup$ Sep 22, 2017 at 11:45

1 Answer 1

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Close, you have calculated $\mathsf P(D_1,D_2)= \mathsf P(D_2\mid A)\,\mathsf P(D_1\mid A)\,\mathsf P(A)+\mathsf P(D_2\mid B)\,\mathsf P(D_1\mid B)\,\mathsf P(B)$

That is not what you seek, but you are almost there.   You have also done the other half of the work, so we will just put it together.

$$\mathsf P(D_2\mid D_1)~{=\dfrac{\mathsf P(D_1,D_2)}{\mathsf P(D_1)}\\ = \dfrac{\mathsf P(D_2\mid A)\mathsf P(D_1\mid A)\mathsf P(A)+\mathsf P(D_2\mid B)\mathsf P(D_1\mid B)\mathsf P(B)}{\mathsf P(D_1\mid A)\mathsf P(A)+\mathsf P(D_1\mid B)\mathsf P(B)}\\=\dfrac{(0.05^2+0.01^2)/2}{(0.05+0.01)/2}\\= \dfrac{0.0013}{0.03}\\=0.04\dot3}$$


Reality check.   $\mathsf P(D_2\mid A)=0.05$ and $\mathsf P(D_2\mid B)=0.01$.   We should find that $\mathsf P(D_2\mid ``\text{some evidence for which source''})$ lies somewhere between these values, and the evidence that the first item was defective is suggestive that the items are more likely to be from source $A$.

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  • $\begingroup$ Can you explain better your notation? you have written $P(D_1, D_2)$, but me $P(D_2|D_1)$, is it the same, or am I wrong? $\endgroup$
    – JB-Franco
    Sep 22, 2017 at 18:23
  • $\begingroup$ and also, in $P(D_2|D_1)$, do you apply Bayes' theorem? But, why $P(D_1, D_2)$ isn't contained within $P(D_1)$? $\endgroup$
    – JB-Franco
    Sep 22, 2017 at 18:35
  • $\begingroup$ Use the definition of conditional probability: $\mathsf P(D_2\mid D_1)=\dfrac{\mathsf P(D_1, D_2)}{\mathsf P(D_1)}$, where $\mathsf P(D_1, D_2)$ is the joint probability for the events, $D_1$ and $D_2$ ; that is the event that both objects are defective. $\endgroup$ Sep 23, 2017 at 4:32

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