A person draws 2 cards from a well shuffled pack of cards, the cards are replaced after noting their colour. Then another person draws 2 cards after shuffling the pack. The probability that there will be exactly 1 common card is

Options

A. `(25)/(546)`
B. `(50)/(663)`
C. `(25)/(663)`
D. None of these

Answer

To solve the problem of finding the probability that there will be exactly 1 common card when two people draw cards from a well-shuffled pack, we can break it down into steps.

### Step 1: Understand the scenario
A person draws 2 cards from a pack of 52 cards, notes their color, and replaces them. Then, another person draws 2 cards from the same pack after shuffling. We want to find the probability that exactly 1 of the 2 cards drawn by the second person is the same as one of the cards drawn by the first person.

### Step 2: Identify the cases
There are two cases for achieving exactly 1 common card:
1. The first card drawn by the second person is the common card, and the second card is different.
2. The first card drawn by the second person is different, and the second card is the common card.

### Step 3: Calculate the probabilities for the first case
- The probability of drawing a common card first:
- There are 2 common cards (the two drawn by the first person) out of 52 total cards.
- Probability = \( \frac{2}{52} \)

- The probability of drawing a different card second:
- After drawing one common card, there are 51 cards left, and only 1 of them is a common card.
- Therefore, there are 50 cards that are different from the common card.
- Probability = \( \frac{50}{51} \)

### Step 4: Calculate the probabilities for the second case
- The probability of drawing a different card first:
- There are 50 cards that are not common out of 52 total cards.
- Probability = \( \frac{50}{52} \)

- The probability of drawing a common card second:
- After drawing one different card, there are still 51 cards left, and 2 of them are common.
- Probability = \( \frac{2}{51} \)

### Step 5: Combine the probabilities
Now, we can combine the probabilities of both cases:
1. For the first case:
\[ P(\text{Case 1}) = \frac{2}{52} \times \frac{50}{51} \]

2. For the second case:
\[ P(\text{Case 2}) = \frac{50}{52} \times \frac{2}{51} \]

Since both cases are mutually exclusive, we can add their probabilities:
\[ P(\text{Exactly 1 common card}) = P(\text{Case 1}) + P(\text{Case 2}) \]

### Step 6: Calculate the total probability
Substituting the values:
\[ P(\text{Exactly 1 common card}) = \left(\frac{2}{52} \times \frac{50}{51}\right) + \left(\frac{50}{52} \times \frac{2}{51}\right) \]
\[ = 2 \times \left(\frac{50}{52} \times \frac{2}{51}\right) \]
\[ = \frac{100}{52 \times 51} \]

### Step 7: Simplify the expression
Now we simplify:
\[ = \frac{100}{2652} = \frac{25}{663} \]

### Final Answer
Thus, the probability that there will be exactly 1 common card is:
\[ \frac{25}{663} \]