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46. 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.

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alok kumar singh | Contributor-Level 10

4 months ago

46. Let, $E_{1} :$ event of choosing a diamond card

$E_{2} :$ event of choosing a card which is not diamond

$A :$ denote the lost card, out of $52$ cards

We know,

$13$ cards are diamond

$39$ 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 $P (A / E_{1}),$

$P (A / E_{1}) = \frac{{}_{12}{C_{2}}}{{}_{51}{C_{2}}} = \frac{12!}{2! \times 10!} \times \frac{2! \times 49!}{51!}$

$= \frac{11 \times 12}{50 \times 51} = \frac{22}{425}$

By using Baye’s theorem,

$P (E_{1} / A) =$ probability that the lost card is diamond, given that the car

...more

46. Let, $E_{1} :$ event of choosing a diamond card

$E_{2} :$ event of choosing a card which is not diamond

$A :$ denote the lost card, out of $52$ cards

We know,

$13$ cards are diamond

$39$ cards are not diamond

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

$P (E_{2}) = \frac{39}{52} = \frac{3}{4}$

The probability of getting two cards, when one diamond card is lost, is given by $P (A / E_{1}),$

$P (A / E_{1}) = \frac{{}_{12}{C_{2}}}{{}_{51}{C_{2}}} = \frac{12!}{2! \times 10!} \times \frac{2! \times 49!}{51!}$

$= \frac{11 \times 12}{50 \times 51} = \frac{22}{425}$

By using Baye’s theorem,

$P (E_{1} / A) =$ probability that the lost card is diamond, given that the card is lost

$P (E_{1} / A) = \frac{P (E_{1}) . P (A / E_{1})}{P (E_{1}) . P (A / E_{1}) + P (E_{2}) . P (A / E_{2})}$

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

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

$= \frac{11}{2} \times \frac{4}{100} = \frac{11}{50}$

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46. 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.

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