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 :
A.
$$\frac{3}{8}$$
B.
$$\frac{1}{6}$$
C.
$$\frac{3}{5}$$
D.
None of these
Answer :
$$\frac{1}{6}$$
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 $$ = \frac{{{}^8{C_3} \times 5!.}}{{8!}} = \frac{1}{6}.$$
Releted MCQ Question on Statistics and Probability >> Probability
Releted Question 1
Two fair dice are tossed. Let $$x$$ be the event that the first die shows an even number and $$y$$ be the event that the second die shows an odd number. The two events $$x$$ and $$y$$ are:
Two events $$A$$ and $$B$$ have probabilities 0.25 and 0.50 respectively. The probability that both $$A$$ and $$B$$ occur simultaneously is 0.14. Then the probability that neither $$A$$ nor $$B$$ occurs is
The probability that an event $$A$$ happens in one trial of an experiment is 0.4. Three independent trials of the experiment are performed. The probability that the event $$A$$ happens at least once is
If $$A$$ and $$B$$ are two events such that $$P(A) > 0,$$ and $$P\left( B \right) \ne 1,$$ then $$P\left( {\frac{{\overline A }}{{\overline B }}} \right)$$ is equal to
(Here $$\overline A$$ and $$\overline B$$ are complements of $$A$$ and $$B$$ respectively).
A.
$$1 - P\left( {\frac{A}{B}} \right)$$
B.
$$1 - P\left( {\frac{{\overline A }}{B}} \right)$$
C.
$$\frac{{1 - P\left( {A \cup B} \right)}}{{P\left( {\overline B } \right)}}$$
D.
$$\frac{{P\left( {\overline A } \right)}}{{P\left( {\overline B } \right)}}$$