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 heart then the probability of the missing card to be a heart is:

A card from a pack of 52 cards is lost. From the remaining 51 cards, two cards are drawn and both are hearts. Using conditional probability (Bayes' Theorem) to find the probability that the lost card was a heart.

The lost card can either be a heart (13 hearts in the deck) or not a heart (39 non-hearts in the deck).

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If the lost card is a heart:

Hearts remaining in deck = 13 - 1 = 12 hearts

Total cards remaining = 51 cards

$P(\text{2 hearts drawn | lost card is heart}) = \frac{12}{51} \times \frac{11}{50}$

$P(\text{2 hearts drawn | lost card is heart}) = \frac{132}{2550}$

$P(\text{lost card is heart}) = \frac{13}{52}$

$P(\text{lost card is heart}) = \frac{1}{4}$

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If the lost card is not a heart:

Hearts remaining in deck = 13 hearts

Total cards remaining = 51 cards

$P(\text{2 hearts drawn | lost card is not heart}) = \frac{13}{51} \times \frac{12}{50}$

$P(\text{2 hearts drawn | lost card is not heart}) = \frac{156}{2550}$

$P(\text{lost card is not heart}) = \frac{39}{52}$

$P(\text{lost card is not heart}) = \frac{3}{4}$

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Using Bayes' Theorem:

$P(\text{lost card is heart | 2 hearts drawn}) = \frac{P(\text{Case 1})}{P(\text{Case 1}) + P(\text{Case 2})}$

For Case 1:

$\frac{132}{2550} \times \frac{1}{4} = \frac{132}{10200}$

For Case 2:

$\frac{156}{2550} \times \frac{3}{4} = \frac{468}{10200}$

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$P = \frac{132}{10200} \div \left(\frac{132}{10200} + \frac{468}{10200}\right)$

$P = \frac{132}{132 + 468}$

$P = \frac{132}{600}$

$P = \frac{11}{50}$

Therefore, the probability of the missing card being a heart is $\frac{11}{50}$.