Two cards are drawn from a pack of 52 cards. What is the probability that either both are black or both are jacks? a) 1/2 b) 1/321 c) 325/1326 d) 55/221

Options

A. 1/2
B. 1/321
C. 325/1326
D. 55/221

Answer

To solve the problem of finding the probability that either both drawn cards are black or both are jacks, we will break it down step by step.

### Step 1: Define Events
Let:
- Event A: Both cards drawn are black.
- Event B: Both cards drawn are jacks.

We need to find the probability of the union of these two events, \( P(A \cup B) \), which can be expressed as:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]

### Step 2: Calculate \( P(A) \)
To find \( P(A) \):
1. The total number of black cards in a deck is 26 (13 spades and 13 clubs).
2. The probability of drawing the first black card is \( \frac{26}{52} \).
3. After drawing one black card, there are now 25 black cards left and 51 total cards remaining. Thus, the probability of drawing a second black card is \( \frac{25}{51} \).

So, the probability \( P(A) \) is:
\[ P(A) = \frac{26}{52} \times \frac{25}{51} = \frac{1}{2} \times \frac{25}{51} = \frac{25}{102} \]

### Step 3: Calculate \( P(B) \)
To find \( P(B) \):
1. The total number of jacks in a deck is 4.
2. The probability of drawing the first jack is \( \frac{4}{52} \).
3. After drawing one jack, there are now 3 jacks left and 51 total cards remaining. Thus, the probability of drawing a second jack is \( \frac{3}{51} \).

So, the probability \( P(B) \) is:
\[ P(B) = \frac{4}{52} \times \frac{3}{51} = \frac{1}{13} \times \frac{3}{51} = \frac{3}{663} = \frac{1}{221} \]

### Step 4: Calculate \( P(A \cap B) \)
To find \( P(A \cap B) \):
- The intersection \( A \cap B \) means both cards are black jacks. There are 2 black jacks (the jack of spades and the jack of clubs).
1. The probability of drawing the first black jack is \( \frac{2}{52} \).
2. After drawing one black jack, there is 1 black jack left and 51 total cards remaining. Thus, the probability of drawing the second black jack is \( \frac{1}{51} \).

So, the probability \( P(A \cap B) \) is:
\[ P(A \cap B) = \frac{2}{52} \times \frac{1}{51} = \frac{1}{26} \times \frac{1}{51} = \frac{1}{1326} \]

### Step 5: Combine the Probabilities
Now we can substitute these values back into the formula for \( P(A \cup B) \):
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
\[ P(A \cup B) = \frac{25}{102} + \frac{1}{221} - \frac{1}{1326} \]

### Step 6: Find a Common Denominator
The least common multiple of the denominators (102, 221, and 1326) is 1326. We will convert each fraction:
1. \( \frac{25}{102} = \frac{25 \times 13}{1326} = \frac{325}{1326} \)
2. \( \frac{1}{221} = \frac{1 \times 6}{1326} = \frac{6}{1326} \)
3. \( \frac{1}{1326} = \frac{1}{1326} \)

Now we can combine:
\[ P(A \cup B) = \frac{325}{1326} + \frac{6}{1326} - \frac{1}{1326} = \frac{325 + 6 - 1}{1326} = \frac{330}{1326} \]

### Step 7: Simplify the Result
Now we simplify \( \frac{330}{1326} \):
\[ \frac{330 \div 6}{1326 \div 6} = \frac{55}{221} \]

### Final Answer
Thus, the probability that either both cards drawn are black or both are jacks is:
\[ \boxed{\frac{55}{221}} \]