There are two die A and B both having six faces. Die A has three faces marked with 1, two faces marked with 2, and one face marked with 3. Die B has one face marked with 1, two faces marked with 2, and three faces marked with 3. Both dices are thrown randomly once. If E be the event of getting sum of the numbers appearing on top faces equal to x and let P(E) be the probability of event E, then P(E) is minimum when x equals to

Options

A. 3
B. 4
C. 5
D. 6

Answer

To solve the problem, we need to find the probability \( P(E) \) of the event \( E \) where the sum of the numbers appearing on the top faces of two dice (A and B) equals \( x \). We will calculate the probabilities for possible sums \( x \) and determine which sum yields the minimum probability.

### Step 1: Identify the faces of each die
Die A has:
- 3 faces marked with 1
- 2 faces marked with 2
- 1 face marked with 3

Die B has:
- 1 face marked with 1
- 2 faces marked with 2
- 3 faces marked with 3

### Step 2: Determine the total outcomes
Each die has 6 faces, so the total number of outcomes when both dice are thrown is:
\[ 6 \times 6 = 36 \]

### Step 3: Calculate probabilities for each possible sum \( x \)

#### Sum \( x = 2 \)
The only combination is (1, 1):
- Probability = \( P(1, 1) = P(A=1) \times P(B=1) = \frac{3}{6} \times \frac{1}{6} = \frac{3}{36} = \frac{1}{12} \)

#### Sum \( x = 3 \)
Possible combinations are (1, 2) and (2, 1):
- Probability = \( P(1, 2) + P(2, 1) = P(A=1) \times P(B=2) + P(A=2) \times P(B=1) \)
- \( = \frac{3}{6} \times \frac{2}{6} + \frac{2}{6} \times \frac{1}{6} = \frac{6}{36} + \frac{2}{36} = \frac{8}{36} = \frac{2}{9} \)

#### Sum \( x = 4 \)
Possible combinations are (1, 3), (2, 2), and (3, 1):
- Probability = \( P(1, 3) + P(2, 2) + P(3, 1) \)
- \( = P(A=1) \times P(B=3) + P(A=2) \times P(B=2) + P(A=3) \times P(B=1) \)
- \( = \frac{3}{6} \times \frac{3}{6} + \frac{2}{6} \times \frac{2}{6} + \frac{1}{6} \times \frac{1}{6} \)
- \( = \frac{9}{36} + \frac{4}{36} + \frac{1}{36} = \frac{14}{36} = \frac{7}{18} \)

#### Sum \( x = 5 \)
Possible combinations are (2, 3) and (3, 2):
- Probability = \( P(2, 3) + P(3, 2) \)
- \( = P(A=2) \times P(B=3) + P(A=3) \times P(B=2) \)
- \( = \frac{2}{6} \times \frac{3}{6} + \frac{1}{6} \times \frac{2}{6} \)
- \( = \frac{6}{36} + \frac{2}{36} = \frac{8}{36} = \frac{2}{9} \)

#### Sum \( x = 6 \)
The only combination is (3, 3):
- Probability = \( P(3, 3) = P(A=3) \times P(B=3) = \frac{1}{6} \times \frac{3}{6} = \frac{3}{36} = \frac{1}{12} \)

### Step 4: Compare probabilities
Now we summarize the probabilities for each sum:
- \( P(E) \) for \( x = 2 \): \( \frac{1}{12} \)
- \( P(E) \) for \( x = 3 \): \( \frac{2}{9} \)
- \( P(E) \) for \( x = 4 \): \( \frac{7}{18} \)
- \( P(E) \) for \( x = 5 \): \( \frac{2}{9} \)
- \( P(E) \) for \( x = 6 \): \( \frac{1}{12} \)

### Step 5: Identify the minimum probability
From the probabilities calculated, we see that:
- The minimum probability occurs at \( x = 2 \) and \( x = 6 \) with \( P(E) = \frac{1}{12} \).

### Conclusion
The event \( E \) has minimum probability when \( x \) equals 2 or 6. However, the question asks for the minimum probability among the options provided (3, 4, 5, 6). The minimum probability occurs at \( x = 4 \) with \( P(E) = \frac{7}{18} \).

Thus, the answer is:
**The minimum probability occurs when \( x = 4 \).**