Question

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

• A. $\frac{1}{3}$
• B. $\frac{1}{2}$
• C. $\frac{2}{3}$
• D. $\frac{1}{6}$

Answer 1

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

The first speaker can be any of A, B or C in $3$ ways.
Then the second speaker can be any of the remaining $2$ persons in $2$ ways.
Lastly, the third speaker can be any of the remaining $1$ person in $1$ way.
So, total number of ways $=3\times 2\times 1=6$ ways
And in the required arrangement where A speaks before B and B speaks before C, is done by only the given $1$ mentioned way.
So, required probability $=\frac{Requiredways}{Totalnumberofways}=\frac{1}{6}$