Four die are thrown simultaneously. The probability that 4 and 3 appear on two of the die given that 5 and 6 have appeared on other two die is a) 1/6 b) 1/36 c) 12/151 d) none of these

Options

A. `1//6`
B. `1//36`
C. `12//151`
D. none of these

Explanation

To solve the problem, we need to find the probability that the numbers 4 and 3 appear on two of the dice given that the numbers 5 and 6 have appeared on the other two dice.

### Step-by-step Solution:

1. **Understanding the Problem**:
- We have 4 dice being thrown.
- We know that two of the dice show the numbers 5 and 6.
- We need to find the probability that the remaining two dice show the numbers 4 and 3.

2. **Identifying the Favorable Outcomes**:
- The two remaining dice can show either (4, 3) or (3, 4).
- Therefore, there are 2 favorable outcomes: (4, 3) and (3, 4).

3. **Calculating the Total Outcomes for the Remaining Dice**:
- Each die has 6 faces, so the total number of outcomes for the two remaining dice is \(6 \times 6 = 36\).

4. **Calculating the Probability**:
- The probability \(P\) that the remaining two dice show 4 and 3 is given by the formula:
\[ P = \frac{\text{Number of Favorable Outcomes}}{\text{Total Outcomes}} \]
- Substituting the values we have:
\[ P = \frac{2}{36} = \frac{1}{18} \]

5. **Conclusion**:
- Since \(\frac{1}{18}\) is not one of the options provided (1/6, 1/36, 12/151, none of these), the correct answer is **none of these**.