# Probability questions of watch problem

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.

• 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? – Ross Millikan May 15 '17 at 5:04
• I found both same answer – S.A May 15 '17 at 5:17
• 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. – S.A May 15 '17 at 5:24
• It sounds to me like the two are guaranteed to come from the same factory – Ross Millikan May 15 '17 at 5:26
• Indian statistical institute – S.A May 15 '17 at 7:46

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 $$P=\frac{78805/80000}{397/400}=\frac{15761}{15880}$$