Four persons are chosen at random from a group of 3 men, 2 women and 4 children. The chance that exactly 2 of them are children, is-
A. $$\frac{{1}}{{9}}$$
B. $$\frac{{1}}{{5}}$$
C. $$\frac{{1}}{{12}}$$
D. $$\frac{{10}}{{21}}$$
E. None of thee
Answer: Option D
Solution(By Examveda Team)
n(S) = number of ways of choosing 4 persons out of 9$$ = {}^9\mathop C\nolimits_4 $$ $$ = \frac{{9 \times 8 \times 7 \times 6}}{{4 \times 3 \times 2 \times 1}}$$ = 126
n(E) = number of ways of choosing 2 children out of 4 and 2 persons out of (3 + 2) personal
n(E) $$ = \left( {{}^4\mathop C\nolimits_2 \times {}^5\mathop C\nolimits_2 } \right)$$ $$ = \left( {\frac{{4 \times 3}}{{2 \times 1}} \times \frac{{5 \times 4}}{{2 \times 1}}} \right)$$ = 60
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{60}}{{126}} = \frac{{10}}{{21}}$$
Related Questions on Probability
A. $$\frac{{1}}{{2}}$$
B. $$\frac{{2}}{{5}}$$
C. $$\frac{{8}}{{15}}$$
D. $$\frac{{9}}{{20}}$$
A. $$\frac{{10}}{{21}}$$
B. $$\frac{{11}}{{21}}$$
C. $$\frac{{2}}{{7}}$$
D. $$\frac{{5}}{{7}}$$
A. $$\frac{{1}}{{3}}$$
B. $$\frac{{3}}{{4}}$$
C. $$\frac{{7}}{{19}}$$
D. $$\frac{{8}}{{21}}$$
E. $$\frac{{9}}{{21}}$$
What is the probability of getting a sum 9 from two throws of a dice?
A. $$\frac{{1}}{{6}}$$
B. $$\frac{{1}}{{8}}$$
C. $$\frac{{1}}{{9}}$$
D. $$\frac{{1}}{{12}}$$
Join The Discussion