Q.Two cards are drawn from a well shuffled deck of $52$ playing cards with replacement. The probability that both cards are queens is
(A) $\dfrac{1}{13} \times \dfrac{1}{13}$
(B) $\dfrac{1}{13} + \dfrac{1}{13}$
(C) $\dfrac{1}{13} \times \dfrac{1}{17}$
(D) $\dfrac{1}{13} \times \dfrac{4}{51}$

Concept: Conditional Probability (with replacement, events are independent).

Step 1: In a standard deck of 52 cards, there are 4 queens.

Probability of drawing a queen in one draw:

$\frac{4}{52} = \frac{1}{13}$.

Step 2: Since the card is replaced, the deck is reset to 52 cards for the second draw. The two draws are independent events. …

The key idea here is conditional probability with independence. When we draw with replacement, the outcome of the first card does not affect the second — the deck is always a full 52 cards. That means the two events are independent, and the joint probability is simply the product of the individual probabilities.

Let’s walk through it step by step.

1. Single-draw probability of a queen A standard deck has 4 queens out of 52 cards. So the probability of drawing a queen on any one draw is

2. Understanding “with replacement”

   After the first card is drawn, it is put back into the deck, and the deck is reshuffled. This means the second draw is from the same full deck of 52 cards. The two draws are independent — knowing the first card’s outcome gives no information about the second.

3. Probability of both being queens

   For independent events, the probability that both occur is the product of their individual probabilities:

4. Matching the options …

Method: Repeated draws with replacement are independent

Use this whenever an item is drawn, returned, and drawn again: the population resets, so each draw has the same probability and the draws are independent.

Steps

Step 1: Spot the phrase "with replacement".

Replacement means the composition of the deck/box is identical before every draw — that is the signal for independence.

Step 2: Find the single-draw probability.

Compute $P(\text{target})$ once from the full population, e.g. $\dfrac{4}{52}=\dfrac{1}{13}$ for a queen. …

Common Mistakes

Mistake 1: Using without-replacement values, $\tfrac{4}{52}\times\tfrac{3}{51}$.

Why it's wrong: the cards are drawn with replacement, so the deck is full ($52$ cards, $4$ queens) on both draws; the $\tfrac{3}{51}$ and $\tfrac{4}{51}$ figures belong to the without-replacement case (distractors C and D). Correct approach: the second draw is still $\tfrac{1}{13}$, so $\tfrac1{13}\times\tfrac1{13}$. …