Question

There are two packs of 52 playing cards. All the four aces from the pack A are removed whereas from the pack B, one ace, one king, one queen and one jack is removed. One of these two packs is selected randomly and two cards are drawn simultaneously from it, and found to be a pair (i.e. both have the same rank e.g two 9’s or two king etc). The probability that the pack A was selected is
A) $\dfrac{12}{23}$
B) $\dfrac{11}{23}$
C) $\dfrac{10}{23}$
D) None of these

Answer 1

This is a question of conditional probability. We will use Bayes theorem to find conditional probability. Choosing of one of the two packs is random and so can be considered as equally probable. We will then find the probability of the event and substitute values in the theorem.

Complete Step by Step Solution:
Bayes Theorem
$\Rightarrow P\left( {B}/{A}\; \right)=\dfrac{P\left( B \right)}{P\left( A \right)}\times P\left( {A}/{B}\; \right)$
A and B are events with probability $P\left( A \right)$ and $P\left( B \right)$ respectively.
$P\left( {B}/{A}\; \right)$ indicates the probability of event B when event A has happened.
In our question
A is the event of getting a pair.
B is the event of choosing pack A.
$P\left( A \right)$ is the probability of getting a pair from both the packs selecting one of them.
This can be done by solving for each case,
Probability of choosing both the packs is the same.
Hence we can take the probabilities to be $\dfrac{1}{2}$ .
Probability of finding a pair is the ratio of total number of pairs available to the sample space of 2 card sets.
In pack A there are no aces and so the pairs can be formed by 4 cards each of only 12 ranks.
In pack B there are all ranks but one card is missing from king, queen, jack and aces. So pairs can be formed by 4 cards of 9 ranks and 3 cards each of the remaining 4 ranks.
Solving accordingly,
$\Rightarrow P\left( A \right)=\dfrac{1}{2}\times \dfrac{12\times {}^{4}{{C}_{2}}}{{}^{48}{{C}_{2}}}+\dfrac{1}{2}\times \dfrac{9\times {}^{4}{{C}_{2}}+4\times {}^{3}{{C}_{2}}}{{}^{48}{{C}_{2}}}$
Here
$\Rightarrow {}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
Exclamation mark indicates factorial function.
As discussed above,
$\Rightarrow P\left( B \right)=\dfrac{1}{2}$
Now for $P\left( {A}/{B}\; \right)$,
Since we have chosen pack A,
we will consider 48 cards with no aces and so only 12 ranks.
$\Rightarrow P\left( {A}/{B}\; \right)=\dfrac{12\times {}^{4}{{C}_{2}}}{{}^{48}{{C}_{2}}}$
Substituting all the values in Bayes Formula
$\Rightarrow P\left( {B}/{A}\; \right)=\dfrac{\dfrac{1}{2}\times \dfrac{12\times {}^{4}{{C}_{2}}}{{}^{48}{{C}_{2}}}}{\dfrac{1}{2}\times \dfrac{12\times {}^{4}{{C}_{2}}}{{}^{48}{{C}_{2}}}+\dfrac{1}{2}\times \dfrac{9\times {}^{4}{{C}_{2}}+4\times {}^{3}{{C}_{2}}}{{}^{48}{{C}_{2}}}}$
Further solving,
$\Rightarrow P\left( {B}/{A}\; \right)=\dfrac{12\times {}^{4}{{C}_{2}}}{12\times {}^{4}{{C}_{2}}+9\times {}^{4}{{C}_{2}}+4\times {}^{3}{{C}_{2}}}$
Calculating this
$\Rightarrow P\left( {B}/{A}\; \right)=\dfrac{12}{12+9+2}$
$\Rightarrow P\left( {B}/{A}\; \right)=\dfrac{12}{23}$

Hence the answer is option A.

Note:
You should apply Bayes theorem only if solving the conditional probability for the reverse case is easy to solve. For example, $P\left( {A}/{B}\; \right)$ was easy to solve and so we simplified using the theorem. You can also find the same for pack B similarly by subtracting $P\left( {B}/{A}\; \right)$ from 1 because both the events are mutually exhaustive.