Probability Question 65
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:
$ 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)} $ .
BETA
