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Factory A produces 1 bad watch in 100 and factory B produces 1 bad watch in 200. You are given two watches from one of the factories (chosen with equal probability) and you don't know which one.

  1. What is the probability that the second watch works?
  2. Given that the first watch works, what is the probability that the second watch works?

According to me, the answer should be $\frac{397}{800}$ for both.

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    $\begingroup$ Which part is your answer for? If you want confirmation, you should explain how you got it. It is much easier to check answers than to generate them. What is your question? $\endgroup$
    – Ross Millikan
    May 15, 2017 at 5:04
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    $\begingroup$ I found both same answer $\endgroup$
    – S.A
    May 15, 2017 at 5:17
  • $\begingroup$ I took M be the event for that both watches chosen from same factory , then P (M)= 1/4 , Again if we consider H1 and H2 are events that 1st or 2nd watch be bad . Then P (H2` int. M)= 1/4 (99/100 +199/200)= 397/800 . And for 2nd case since the events H1 and H2 are Independent Thus H2` remains same. $\endgroup$
    – S.A
    May 15, 2017 at 5:24
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    $\begingroup$ It sounds to me like the two are guaranteed to come from the same factory $\endgroup$
    – Ross Millikan
    May 15, 2017 at 5:26
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    $\begingroup$ Indian statistical institute $\endgroup$
    – S.A
    May 15, 2017 at 7:46

1 Answer 1

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For part 1, since the watches are independently picked: $$P=\frac12\cdot\frac{99}{100}+\frac12\cdot\frac{199}{200}=\frac{397}{400}$$ which is not $\frac{397}{800}$.

For part 2, use the law of total probability. The probability that both watches work is $$\frac12\cdot\frac{99^2}{100^2}+\frac12\cdot\frac{199^2}{200^2}=\frac{78805}{80000}$$ The probability of the first watch working is the same as the answer to part 1. Therefore $$P=\frac{78805/80000}{397/400}=\frac{15761}{15880}$$

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  • $\begingroup$ (Chosen with equal probably )is not given on question .someone edited it here. So I can't agree about 1/2 , I think it will be1/4. @part 1 . $\endgroup$
    – S.A
    May 15, 2017 at 7:59
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    $\begingroup$ @S.A If I hadn't edited that in, how could I answer the question? $\endgroup$
    – Parcly Taxel
    May 15, 2017 at 8:00
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    $\begingroup$ Hummmm u r right. $\endgroup$
    – S.A
    May 15, 2017 at 8:02

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