Statistics And Probability Question 236
Question: If m rupee coins and n ten paise coins are placed in a line, then the probability that the extreme coins are ten paise coins is
Options:
A) $ ^{m+n}C_{m}/n^{m} $
B) $ \frac{n,(n-1)}{(m+n),(m+n-1)} $
C) $ ^{m+n}P_{m}/m^{n} $
D) $ ^{m+n}P_{n}/n^{m} $
Show Answer
Answer:
Correct Answer: B
Solution:
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$ m $ rupee coins and $ n $ ten paise coins can be placed in a line in $ \frac{(m+n)!}{m!n!} $ ways. If the extreme coins are ten paise coins, then the remaining $ n-2 $ ten paise coins and $ m $ one rupee coins can be arragned in a line in $ \frac{(m+n-2)!}{m!(n-2)!} $ ways. Hence the required probability $ =\frac{\frac{(m+n-2)!}{m!(n-2)!}}{\frac{(m+n)!}{m!n!}}=\frac{n(n-1)}{(m+n)(m+n-1)} $ .
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