Probability Question 371
Question: Three persons A, B and C are to speak at a function along with five others. If they all speak in random order, the probability that A speaks before B and B speaks before C, is
Options:
A) $ \frac{3}{8} $
B) $ \frac{1}{6} $
C) $ \frac{3}{5} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
The total number of ways in which 8 persons can speak is $ ^{8}P_8=8! $ the number of ways in which, A, B and C can be arranged in the specified speaking order is $ ^{8}C_3. $
There are 5!
Ways in which the other five can speak.
So, favourable number of ways is $ ^{8}C_3\times 5! $
Hence, required probability $ =\frac{^{8}C_3\times 5!}{8!}=\frac{1}{6}. $
BETA
