Question

If there are two red and blue dice. Find the probability when the sum of the dice is a prime number, where the number on red is more than blue.

• A. 1/12
• B. 5/36
• C. 7/36
• D. 1/9

Hint:

Use the prime sums $3,5,7,11$. Because the dice have colours, keep ordered pairs and impose red greater than blue before dividing by $36$.

Answer 1

The Correct Option is B

The possible prime numbers that can be the sum of two dice are 3, 5, 7, and 11. 
For each of these sums: 
- 3: (1,2)
- 5: (1,4), (2,3)
- 7: (1,6), (2,5), (3,4)
- 11: (5,6)
Out of these, only the pairs where the number on the red die is greater than the blue die are considered. The favorable pairs are: 
- (2,1), (3,2), (6,1), (5,2), (6,5)
Thus, the probability is:
 

\[P = \frac{5}{36}\]

Answer 2

Concept:

• The dice are distinguishable, so there are $36$ ordered outcomes.
• For an odd prime sum, the two dice cannot be equal; exactly half of the valid ordered pairs have red greater than blue.

Step 1: Count outcomes with a prime sum.
Possible prime sums are $3,5,7,11$. Their ordered-pair counts are $2,4,6,2$, giving $14$ outcomes.

Step 2: Apply the colour-order condition.
Swapping the red and blue values pairs every valid outcome with one having the opposite inequality. Therefore, $14/2=7$ outcomes have red greater than blue.

Step 3: Divide by the sample-space size.
$P=\dfrac{7}{36}$.

Final Answer: $\dfrac{7}{36}$, option C