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I ran into a problem that seems strange to me.

Two companies A,B produce a device that with probability $0.05$ and $0.01$ are broken. if we buy two devices produced by one company with equal probability and the first device be broken, what is the probability of the second device be broken?

Who can show me how my TA reached to $13/300$ ?

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  • $\begingroup$ So, you pick a random company with even probability, buy two devices, and the first you test is broken? $\endgroup$ – Thomas Andrews Apr 15 '15 at 19:13
  • $\begingroup$ yes @ThomasAndrews okey $\endgroup$ – Zyxef Kos Apr 15 '15 at 19:19
  • $\begingroup$ @ThomasAndrews are you agree with this answer? I think something is wrong ?! $\endgroup$ – Zyxef Kos Apr 15 '15 at 19:40
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Denote by $K$ the event that the first checked device is broken. Then $$P(K)=P(A) P(K|A)+P(B)P(K|B)={1\over2}\cdot0.05+{1\over2}\cdot0.01=0.03\ .$$ What we need to know are the conditional probabilities $P(A|K)$, $P(B|K)$. Now $$P(K)P(A|K)=P(A\cap K)=P(A)(P(K|A)\ ,$$ which implies $$P(A|K)={P(A)P(K|A)\over P(K)}={{1\over2}\cdot0.05\over0.03}={5\over6}\ ,$$ and similarly $$P(B|K)={1\over6}\ .$$ Therefore the event $K'$ that the second checked device is broken has probability $$P(K')=P(A|K)P(K|A)+P(B|K)P(K|B)={5\over6}\cdot 0.05+{1\over6}\cdot0.01={13\over300}\ .$$

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A breaks 5 times as often as B. So for every 6 broken devices, 5 times it will be A, and 1 time it will be B.

The broken device was created by A with a $\frac{5}{6}$ probability. Multiplying $\frac{5}{6}$ by the chance A will break the next device, $\frac{1}{20}$, gives $\frac{1}{24}$.

Now the broken device has a 1/6 probability of being created by B. So multiplying by the 1/100 probability B will break the next device gives 1/600. Adding, $\frac{1}{600}+\frac{1}{24}$=$\boxed{\frac{13}{300}}$

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  • $\begingroup$ How you get 1/20 ? $\endgroup$ – Zyxef Kos Apr 16 '15 at 6:01
  • $\begingroup$ A has a 5% chance of breaking its item, which is equal to 5/100 or 1/20 $\endgroup$ – user156213 Apr 16 '15 at 17:48

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