A card from a pack of $52$ cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. If the probability of the lost card is a diamond is $p$ enter $100 p$

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Solution

Let $E_{1}$ and $E_{2}$ be the respective events of choosing a diamond card and a card which is not diamond.
Let $A$ denote the lost card.
Out of $52$ cards, $13$ cards are diamond and $39$ cards are not diamond.
$∴ P (E_{1}) = \frac{13}{52} = \frac{1}{4}$
$P (E_{2}) = \frac{39}{52} = \frac{3}{4}$
When one diamond card is lost, there are $12$ diamond cards out of $51$ cards.
Two cards can be drawn out of $12$ diamond cards in $^{12} C_{2}$ ways.
Similarly, $2$ diamond cards can be drawn out of $51$ cards in $^{51} C_{2}$ ways. The probability of getting two cards, when one diamond card is lost, is given by $P (A | E_{1})$ .
$P (A | E_{1}) = \frac{^{12} C_{2}}{^{51} C_{2}} = \frac{12!}{2! \times 10!} \times \frac{2! \times 49!}{51!} = \frac{11 \times 12}{50 \times 51} = \frac{22}{425}$
When the lost card is not a diamond, there are $13$ diamond cards out of $51$ cards.
Two cards can be drawn out of $13$ diamond cards in $^{13} C_{2}$ ways whereas $2$ cards can be drawn out of $51$ cards in $^{51} C_{2}$ ways.
The probability of getting two cards, when one card is lost which is not diamond, is given by $P (A | E_{2})$ .
$P (A | E_{2}) = \frac{^{13} C_{2}}{^{51} C_{2}} = \frac{13!}{2! \times 11!} \times \frac{2! \times 49!}{51!} = \frac{12 \times 13}{50 \times 51} = \frac{26}{425}$
The probability that the lost card is diamond is given by $P (E_{1} | A)$ .
By using Baye's theorem, we obtain
$P (E_{1} | A) = \frac{P (E_{1}) \cdot P (A | E_{1})}{P (E_{1}) \cdot P (A | E_{1}) + P (E_{2}) \cdot P (A | E_{2})}$
$= \frac{\frac{1}{4} \cdot \frac{22}{425}}{\frac{1}{4} \cdot \frac{22}{425} + \frac{3}{4} \cdot \frac{26}{425}}$
$= \frac{\frac{1}{425} (\frac{22}{4})}{\frac{1}{425} (\frac{22}{4} + \frac{26 \times 3}{4})}$
$= \frac{\frac{11}{2}}{25}$
$= \frac{11}{50}$

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