Question

In a room there are six couples. Out of them if $4$ people are selected at random, the probability that they may be couples is

• A. $\frac{4}{33}$
• B. $\frac{2}{33}$
• C. $\frac{1}{33}$
• D. $\frac{3}{33}$

Answer 1

Answer: (C)

$\frac{1}{33}$

$4$ people so $2$ couple
Favorable cases: ${}^6{C}_2$
Total cases: ${}^{12}{C}_4$
Probability: $\frac{{}^6{C}_2}{{}^{12}{C}_4}$
$=\frac{1}{33}$