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 drawn two cards at random and says, ‘ I hold at least one ace’. The probability that he holds two aces are: -
(a) \[\dfrac{2}{8}\]
(b) \[\dfrac{4}{9}\]
(c) \[\dfrac{2}{3}\]
(d) \[\dfrac{1}{9}\]

Answer 1

Hint: Find the probability of drawing two aces from the 16 shuffled cards using the formula: - \[\dfrac{{}^{4}{{C}_{2}}}{{}^{16}{{C}_{2}}}\], here 16 is the total number of cards and 4 is the total number of aces. Now, find the probability to draw one ace and one more card other than ace from the remaining 12 cards. The required expression will be \[\dfrac{{}^{4}{{C}_{1}}\times {}^{12}{{C}_{1}}}{{}^{16}{{C}_{2}}}\]. Add the two obtained probabilities to determine the probability of drawing at least one ace. Finally, use conditional probability to determine the probability of holding two aces given that the opponent has at least one ace by dividing \[\dfrac{{}^{4}{{C}_{2}}}{{}^{16}{{C}_{2}}}\] with the sum of \[\dfrac{{}^{4}{{C}_{2}}}{{}^{16}{{C}_{2}}}\] and \[\dfrac{{}^{4}{{C}_{1}}\times {}^{12}{{C}_{1}}}{{}^{16}{{C}_{2}}}\].

Complete step by step answer:
We have been given that the opponent always tells the truth, that means it is true that he is holding at least one ace. Opponent draws two cards.
\[\Rightarrow \]Probability of drawing one ace and one other card,
\[=\dfrac{{}^{4}{{C}_{1}}\times {}^{12}{{C}_{1}}}{{}^{16}{{C}_{2}}}\], since there are 4 aces and 12 cards other than aces
Also, probability of drawing two aces,
\[=\dfrac{{}^{4}{{C}_{2}}}{{}^{16}{{C}_{2}}}\]
Therefore, probability of drawing at least one ace
= Probability of drawing 1 ace + Probability of drawing two ace.
\[=\dfrac{{}^{4}{{C}_{1}}\times {}^{12}{{C}_{1}}}{{}^{16}{{C}_{2}}}+\dfrac{{}^{4}{{C}_{2}}}{{}^{16}{{C}_{2}}}=\dfrac{\left( {}^{4}{{C}_{1}}\times {}^{12}{{C}_{1}} \right)+{}^{4}{{C}_{2}}}{{}^{16}{{C}_{2}}}\]
Now, we have to find the probability of holding two aces with given condition is, opponent is holding one ace.
\[\Rightarrow \] Required probability
= (Probability of drawing 2 aces) / (Probability of drawing at least 1 ace)
\[\begin{align}
  & =\dfrac{\dfrac{{}^{4}{{C}_{2}}}{{}^{16}{{C}_{2}}}}{\dfrac{\left( {}^{4}{{C}_{1}}\times {}^{12}{{C}_{1}} \right)+{}^{4}{{C}_{2}}}{{}^{16}{{C}_{2}}}} \\
 & =\dfrac{{}^{4}{{C}_{2}}}{{}^{4}{{C}_{1}}\times {}^{12}{{C}_{1}}+{}^{4}{{C}_{2}}} \\
 & =\dfrac{6}{4\times 12+6} \\
 & =\dfrac{6}{54} \\
 & =\dfrac{1}{9} \\
\end{align}\]

So, the correct answer is “Option D”.

Note: One may note that this question cannot be solved by simple probability because it contains condition and hence we need the concept of conditional probability. You may see that when we have drawn one ace then we have used the expression \[{}^{4}{{C}_{1}}\times {}^{12}{{C}_{1}}\], this is because two cards are drawn in which one must be ace and the other must not be ace. So, there are 4 aces and 12 cards other than ace so the other card must be drawn from these 12 cards only.