A card form a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn at random and are found to be both clubs. Find the probability of the lost card being of clubs.

Solution

Let the events E₁, E_2,E₃, E₄ and A be defined as follows:

E₁: Missing card is a diamond

E₂: Missing card is a spade

E₃: Missing card is a ciub

E₄: Missing card is a heart

A: Drawing two club cards

P ( E₁ ) = P ( E₂ ) = P ( E₃ ) = P ( E₄ ) = $\frac{1}{4}$

$p (A | E_{1}) = p (A | E_{2}) = p (A | E_{4}) = \frac{13}{51} \times \frac{12}{50} p (A | E_{3}) = \frac{12}{51} \times \frac{11}{50} p (E_{3} | A) = \sum_{i = 1}^{4} \frac{p (E_{3}) p (A | E_{3})}{P (E_{i}) p (A | E_{i})} = \frac{1}{4} \times \frac{12}{51} \times \frac{11}{50} = \frac{1}{4} [\frac{13 x 12 + 13 x 12 + 13 x 12 + 11 x 12}{51 x 50}] = \frac{12 x 11}{3 x 13 x 12 + 12 x 11} = \frac{11}{50}$

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Probability

A card form a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn at random and are found to be both clubs. Find the probability of the lost card being of clubs.

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