Question

Two cards are drawn randomly from a pack of 52 playing cards one after the other with replacement. If A is the event of drawing a face card in the first draw and B is the event of drawing a clubs card in the second draw, then $P\left(\frac{B}{A}\right)=$

• A. $\frac{11}{12}$
• B. $\frac{12}{13}$
• C. $\frac{3}{4}$
• D. $\frac{1}{4}$

Hint:

The keywords "with replacement" are crucial in probability problems involving sequential draws. They almost always signify that the events are independent. For independent events A and B, remember that $P(A|B) = P(A)$ and $P(B|A) = P(B)$.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
The problem asks for a conditional probability, $P(B|A)$, which is the probability of event B occurring given that event A has already occurred. However, the draws are made "with replacement". This implies that the outcome of the first draw does not affect the outcome of the second draw. Therefore, the events A and B are independent.
Step 2: Key Formula or Approach
For two independent events A and B, the conditional probability of B given A is simply the probability of B. \[ P(B|A) = P(B) \] We need to calculate the probability of event B, which is drawing a clubs card in the second draw.
Step 3: Detailed Explanation
Event A: Drawing a face card in the first draw. A standard deck has 52 cards. Face cards are Jack (J), Queen (Q), King (K). There are 4 suits. Number of face cards = $3 \times 4 = 12$. $P(A) = \frac{12}{52} = \frac{3}{13}$. Event B: Drawing a clubs card in the second draw. There are 13 clubs in a deck of 52 cards. $P(B) = \frac{13}{52} = \frac{1}{4}$. The draws are made with replacement. This means after the first card is drawn, it is put back into the deck before the second card is drawn. The deck is restored to its original 52 cards for the second draw. Consequently, the result of the first draw has no influence on the result of the second draw. The events A and B are independent. For independent events, the conditional probability $P(B|A)$ is defined as: \[ P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{P(A)P(B)}{P(A)} = P(B) \] Therefore, we just need the probability of event B. \[ P(B|A) = P(B) = \frac{13}{52} = \frac{1}{4} \] Step 4: Final Answer
Since the draws are with replacement, the events are independent. The probability of drawing a clubs card in the second draw is independent of the first draw's outcome. Therefore, $P(B|A) = P(B) = \frac{13}{52} = \frac{1}{4}$.