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A card from a pack of 52 cards is lost.

Question:

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. Find the probability of the lost card being a diamond.

Solution:

Let E1 and Ebe 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.

$\therefore P\left(E_{1}\right)=\frac{13}{52}=\frac{1}{4}$

$\mathrm{P}\left(\mathrm{E}_{2}\right)=\frac{39}{52}=\frac{3}{4}$

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 ${ }^{12} \mathrm{C}_{2}$ ways.

Similarly, 2 diamond cards can be drawn out of 51 cards in ${ }^{51} \mathrm{C}_{2}$ ways. The probability of getting two cards, when one diamond card is lost, is given by $\mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{1}\right)$.

$\mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{1}\right)=\frac{{ }^{12} C_{2}}{{ }^{51} C_{2}}=\frac{12 !}{2 ! \times 10 !} \times \frac{21 \times 49 !}{51 !}=\frac{11 \times 12}{50 \times 51}=\frac{22}{425}$

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 ${ }^{13} \mathrm{C}_{2}$ ways whereas 2 cards can be drawn out of 51 cards in ${ }^{51} \mathrm{C}_{2}$ ways.

The probability of getting two cards, when one card is lost which is not diamond, is given by $P\left(A \mid E_{2}\right)$.

$\mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{2}\right)=\frac{{ }^{13} C_{2}}{{ }^{51} C_{2}}=\frac{13 !}{2 ! \times 11 !} \times \frac{2 ! \times 49 !}{51 !}=\frac{12 \times 13}{50 \times 51}=\frac{26}{425}$

The probability that the lost card is diamond is given by $P\left(E_{1} \mid A\right)$.

By using Bayes’ theorem, we obtain

$P\left(E_{1} \mid A\right)=\frac{P\left(E_{1}\right) \cdot P\left(A \mid E_{1}\right)}{P\left(E_{1}\right) \cdot P\left(A \mid E_{1}\right)+P\left(E_{2}\right) \cdot P\left(A \mid E_{2}\right)}$

$=\frac{\frac{1}{4} \cdot \frac{22}{425}}{\frac{1}{4} \cdot \frac{22}{425}+\frac{3}{4} \cdot \frac{26}{425}}$

$=\frac{\frac{1}{425}\left(\frac{22}{4}\right)}{\frac{1}{425}\left(\frac{22}{4}+\frac{26 \times 3}{4}\right)}$

$=\frac{11}{2}$

$=\frac{11}{50}$

 

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