CBSE 2024, Set 3

Question

A card from a well shuffled deck of 52 playing cards is lost. From the remaining cards of the pack, a card is drawn at random and is found to be a King. Find the probability of the lost card being a King.

Edvaya · Accepted Answer

The correct answer is Subjective. **Step 1: Define the events and their probabilities**

Let $E_1$ be the event that the lost card is a King.
Let $E_2$ be the event that the lost card is not a King.
Let $A$ be the event that the subsequently drawn card is a King.

Initial probabilities of losing a card from the 52-card deck (which has 4 Kings and 48 non-Kings):
$$P(E_1) = \frac{4}{52} = \frac{1}{13}$$
$$P(E_2) = \frac{48}{52} = \frac{12}{13}$$

**Step 2: Find the conditional probabilities**

If $E_1$ occurs (a King is lost), the deck now has 51 cards with only 3 Kings remaining:
$$P(A|E_1) = \frac{3}{51}$$

If $E_2$ occurs (a non-King is lost), the deck has 51 cards with all 4 Kings still remaining:
$$P(A|E_2) = \frac{4}{51}$$

**Step 3: Apply Bayes' Theorem**

We need to find the probability that the lost card was a King, given that a King was drawn, which is $P(E_1|A)$.

$$P(E_1|A) = \frac{P(E_1) \cdot P(A|E_1)}{P(E_1) \cdot P(A|E_1) + P(E_2) \cdot P(A|E_2)}$$

Substitute the values:
$$P(E_1|A) = \frac{\left(\frac{1}{13}ight) \left(\frac{3}{51}ight)}{\left(\frac{1}{13}ight) \left(\frac{3}{51}ight) + \left(\frac{12}{13}ight) \left(\frac{4}{51}ight)}$$

Cancel the common denominator $(13 	imes 51)$ from all terms:
$$P(E_1|A) = \frac{1 	imes 3}{(1 	imes 3) + (12 	imes 4)}$$
$$P(E_1|A) = \frac{3}{3 + 48} = \frac{3}{51}$$

Simplify the fraction:
$$P(E_1|A) = \frac{1}{17}$$

**Therefore, the probability that the lost card was a King is $\frac{1}{17}$.**

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