$$10$$ different books and $$2$$ different pens are given to $$3$$ boys so that each gets equal number of things. The probability that the same boy does not receive both the pens is :
A.
$$\frac{5}{{11}}$$
B.
$$\frac{7}{{11}}$$
C.
$$\frac{2}{3}$$
D.
$$\frac{6}{{11}}$$
Answer :
$$\frac{6}{{11}}$$
Solution :
$$n\left( S \right) = {}^{12}{C_4} \times {}^8{C_4} \times {}^4{C_4}$$
$$n\left( E \right) = n\left( S \right) - $$ the number of ways in which one boy gets both the pens
$$ = n\left( S \right) - {}^{10}{C_2} \times {}^8{C_4} \times {}^4{C_4} \times \left( {3!} \right)$$
$$\therefore \,P\left( E \right) = 1 - \frac{{{}^{10}{C_2} \times {}^8{C_4} \times {}^4{C_4} \times \left( {3!} \right)}}{{{}^{12}{C_4} \times {}^8{C_4} \times {}^4{C_4}}} = 1 - \frac{6}{{11}} = \frac{5}{{11}}.$$
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.
$$\frac{{P\left( {\overline A } \right)}}{{P\left( {\overline B } \right)}}$$