$$4$$ gentlemen and $$4$$ ladies take seats at random round a table. The probability that they are sitting alternately is :
A.
$$\frac{4}{{35}}$$
B.
$$\frac{1}{{70}}$$
C.
$$\frac{2}{{35}}$$
D.
$$\frac{1}{{35}}$$
Answer :
$$\frac{1}{{35}}$$
Solution :
$$n\left( S \right) = 7!,\,n\left( E \right) = \left( {3!} \right) \times \left( {4!} \right)$$ ($$\because $$ after making $$4$$ gentlemen sit in $$3!$$ ways, $$4$$ ladies can sit in $$4!$$ ways in between the
gentlemen)
$$\therefore \,P\left( E \right) = \frac{{\left( {3!} \right) \times \left( {4!} \right)}}{{7!}} = \frac{6}{{7 \times 6 \times 5}} = \frac{1}{{35}}.$$
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.
$$\frac{{P\left( {\overline A } \right)}}{{P\left( {\overline B } \right)}}$$