Question

Three persons A, B and C are to speak at a function along with 5 other persons. If the persons speak in random order, the probability that A speaks before B and B speaks before C is:

• A. $\frac{3}{8}$
• B. $\frac{1}{6}$
• C. $\frac{3}{5}$
• D. $\frac{4}{15}$

Answer 1

The correct option is B $\frac{1}{6}$

Total ways = 8!
Since the order of A, B and C are fixed, we have to divide this by 3!(number of ways we can arrange A, B and C) to get the number of favorable ways.
Therefor number of favourable ways = $\frac{8!}{3!}$
$\therefore$ Probability =$\frac{1}{3!}=\frac{1}{6}$