Question

Three persons $A,B$ and $C$ speak at a function along with $5$ other persons. If persons speak at random, then the probability that $A$ speaks before $B$ and $B$ speaks before $C$ is

• A. $\frac{1}{6}$
• B. $\frac{1}{4}$
• C. $\frac{2}{5}$
• D. $\frac{1}{8}$

Answer 1

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

Let $E$ be the event where ''$A$ speaks before $B$ and $B$ speaks before $C$''
There are total $8$ persons to speak.
$\therefore n\left(S\right)=8!$

As $A$ speaks before $B$ and $B$ speaks before $C.$
$\therefore A,B$ and $C$ can be arranged in only one order and we have to select three places for $A,B$ and $C$
Remaining persons can be arranged in $5!$ ways.
$\therefore n\left(E\right)={}^8{C}_3\times 1\times 5!$
$\therefore P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{{}^8{C}_3\times 1\times 5!}{8!}=\frac{1}{6}$