To solve this problem, let's analyze the probability of rolling two dice twice and getting a sum of 3 on the first roll and a sum of 4 on the second roll.

Step 1: Find the probability of getting a sum of 3 on the first roll.

From the given table, the probability of getting a sum of 3 is: $P(\text{sum of 3}) = \frac{2}{36}$

Step 2: Find the probability of getting a sum of 4 on the second roll.

Similarly, the probability of getting a sum of 4 is: $P(\text{sum of 4}) = \frac{3}{36}$

Step 3: Determine if these events are independent.

Rolling two dice twice are independent events because the outcome of the first roll does not affect the outcome of the second roll.

Step 4: Calculate the combined probability.

Since the events are independent, the probability of both events occurring (getting a sum of 3 on the first roll and a sum of 4 on the second roll) is the product of the individual probabilities: $P(\text{sum of 3 on first roll and sum of 4 on second roll}) = \left(\frac{2}{36}\right) \times \left(\frac{3}{36}\right) = \frac{6}{1296} = \frac{1}{216}$

Conclusion

The correct answer is: C. The probability is $\frac{2}{36} \times \frac{3}{36}$ because the two events are independent.

Would you like further details or have any other questions?

Here are some related questions for you to consider:

1. What is the probability of rolling a sum of 7 on two consecutive rolls?
2. How would the probability change if the dice were rolled three times instead of two?
3. Can the probabilities of independent events ever exceed 1 when multiplied?
4. How would you calculate the probability of not rolling a sum of 3 or 4 at all?
5. What is the probability of getting the same sum in two consecutive rolls?

Tip: When dealing with independent events, remember that the combined probability is the product of individual probabilities.