Question

All the jacks, queens, kings and aces of a regular 52 cards deck are taken out. The 16 cards are thoroughly shuffled and my opponent, a person who always tells the truth, simultaneously draws two cards at random and says, 'I hold at least one ace'. The probability that he holds two aces is

• A. $2/8$
• B. $4/9$
• C. $2/3$
• D. $1/9$

Answer 1

The correct option is D $1/9$

A pack of deck has $52$ cards. Out of which $16$ cards are taken which includes $4$ jack, $4$ queens, $4$ king and $4$ aces.

$\Rightarrow$ Probability that the man always tells truth $=\frac{1}{2}$

$\Rightarrow P$ (at least one ace) $=P\left(1\right)+P\left(2\right)$

$=\frac{{}^4{C}_1+{}^4{C}_2}{{}^{16}{C}_2}$

$=\frac{1}{12}$

$\Rightarrow$ Here total no. of cards taken i.e, total probability $={}^{16}{C}_2$

Hence, the answer is $\frac{1}{9}.$