Two cards are drawn at random from a pack of 52 cards. The probability that either both are black or both are queen is :

Question

Accepted Answer

To solve the problem of finding the probability that either both cards drawn are black or both are queens, we need to consider the following:

1. **Total number of ways to draw two cards from a deck of 52 cards:**

The total number of ways to draw 2 cards from 52 is given by the combination formula:
   $$
   \binom{52}{2} = \frac{52 \times 51}{2 \times 1} = 1326
   $$

2. **Probability that both cards are black:**

There are 26 black cards in a deck (13 spades and 13 clubs). The number of ways to draw 2 black cards is:
   $$
   \binom{26}{2} = \frac{26 \times 25}{2 \times 1} = 325
   $$

3. **Probability that both cards are queens:**

There are 4 queens in a deck. The number of ways to draw 2 queens is:
   $$
   \binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6
   $$

4. **Probability that both cards are black queens:**

There are 2 black queens (queen of spades and queen of clubs). The number of ways to draw 2 black queens is:
   $$
   \binom{2}{2} = 1
   $$

5. **Using the principle of inclusion-exclusion:**

The probability that either both cards are black or both are queens is given by:
   $$
   P(\text{both black or both queens}) = P(\text{both black}) + P(\text{both queens}) - P(\text{both black queens})
   $$

Substituting the values calculated:
   $$
   = 325 + 6 - 1 = 330
   $$

6. **Probability calculation:**

The probability is then:
   $$
   \frac{330}{1326}
   $$

Thus, the correct answer is Option 3: $\frac{330}{1326}$.

**Explanation for why other options are incorrect:**

- **Option 1: $\frac{325}{1326}$** is incorrect because it only accounts for the probability of both cards being black, without considering the queens.

- **Option 2: $\frac{6}{1326}$** is incorrect because it only accounts for the probability of both cards being queens, without considering the black cards.

- **Option 4: $\frac{5}{1326}$** is incorrect as it does not correspond to any logical combination of the probabilities involved in this scenario.

Therefore, the correct answer, supported by the calculations and logical reasoning, is Option 3: $\frac{330}{1326}$.