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There are two local factories that produce microwaves. Each microwave produced at factory A is defective with probability $.05$, whereas each one produced at factory B is defective with probability $.01$. Suppose you purchase two microwaves that were produced at the same factory, which is equally likely to have been either factory A or factory B. If the first microwave that you check is defective, what is the conditional probability that the other one is also defective?

Ok so below is my attempts to this problem. Obviously what I got is wrong, but I'm confused on what I did that was wrong. I'm trying to figure out what is wrong with my understanding. Note: I didn't know how to include fancy symbols like I see on other post so I just installed a picture instead.

My work: My work

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Use your notations: $D1$ for the event that the first one is defective and $D2$ for the event that the second is defective. You want to find $P(D2|D1)$.

$$P(D2|D1)=\frac{P(D2\cap D1)}{P(D1)}=\frac{P(D2\cap D1\cap A)+P(D2\cap D1\cap B)}{P(D1\cap A)+P(D1\cap B)}=\frac{P(D2\cap D1|A)P(A)+P(D2\cap D1| B)P(B)}{P(D1|A)P(A)+P(D1|B)P(B)}=\frac{0.05^2(0.5)+0.01^2(0.5)}{0.05(0.5)+0.01(0.5)}$$

Compare your writing and try to find out where you made mistakes. For example, your first equation after $P(D2|D1)$ is wrong. You know that $P(D1)=P(D1|A)P(A)+P(D1|B)P(B)$. You just need to treat "$D1\cap D2$" in the numerator as "$D1$" in the denominator.

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  • $\begingroup$ After the first equation, why couldn't I write the numerator as [P(D2/A U D2 /B) U P(D1/A U D2/A)]... Don't I have to find the P(all D2, including from factory A and B) U P(all D1, including from factory A and B)? $\endgroup$ Oct 22 at 4:26
  • $\begingroup$ $P(D_2|A \cup D_2 |B)$ is not a correct expression. It can't be conditioned on two disjoint events, which doesn't make sense. Following the definitions and go step by step. $\endgroup$ Oct 22 at 16:43
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Let’s conduct $20,000$ trials, $10,000$ from each factory. In $600$ of those trials ($500$ from Factory A and $100$ from Factory B), the first microwave you test will be defective. In $26$ of those $600$ trials ($25$ from Factory A and $1$ from Factory B), the second microwave also will be defective. So the conditional probability that the microwaves are both defective is $\frac{26}{600}=\frac{13}{300}$.

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