Question

Two dice are rolled simultaneously. The probability that the sum of the two numbers on the dice is a prime number, is

• A. $\frac{5}{11}$
• B. $\frac{5}{12}$
• C. $\frac{7}{12}$
• D. $\frac{7}{11}$

Hint:

Instead of re-counting combinations during exams, try remembering that out of the 36 total outcomes for two dice, the number of ways to get a prime sum is exactly 15. The simplified fraction $\frac{15}{36} = \frac{5}{12}$ is a very frequent flyer in probability questions!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
When two fair six-sided dice are rolled at the same time, the sum of the appearing numbers can range anywhere from a minimum of 2 to a maximum of 12. We need to calculate the mathematical probability that this resulting sum is a prime number.

Step 2: Key Formula or Approach:
The probability of an event $E$ occurring is given by the standard formula: $$ P(E) = \frac{n(E)}{n(S)} $$ where $n(S)$ represents the total number of possible outcomes in the sample space, and $n(E)$ represents the number of favorable outcomes matching our condition.

Step 3: Detailed Explanation:
First, let's establish the size of the total sample space $S$. Since each die has 6 distinct faces, rolling two dice simultaneously yields: $$ n(S) = 6 \times 6 = 36 $$ Next, let's identify all the possible prime numbers between the minimum sum of 2 and the maximum sum of 12. The prime numbers in this range are 2, 3, 5, 7, and 11. Now, let's systematically list the favorable ordered pairs $(x, y)$ from the two dice that add up to each of these prime numbers:

• Sum = 2: $(1, 1)$ $\rightarrow$ 1 outcome

• Sum = 3: $(1, 2), (2, 1)$ $\rightarrow$ 2 outcomes

• Sum = 5: $(1, 4), (2, 3), (3, 2), (4, 1)$ $\rightarrow$ 4 outcomes

• Sum = 7: $(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)$ $\rightarrow$ 6 outcomes

• Sum = 11: $(5, 6), (6, 5)$ $\rightarrow$ 2 outcomes
Summing these favorable outcomes up to find $n(E)$: $$ n(E) = 1 + 2 + 4 + 6 + 2 = 15 $$ Substituting these values into our probability equation: $$ P(E) = \frac{15}{36} $$ Dividing both the numerator and denominator by their greatest common divisor, 3, simplifies the fraction: $$ P(E) = \frac{15 \div 3}{36 \div 3} = \frac{5}{12} $$

Step 4: Final Answer: The probability that the sum of the two numbers is a prime number is $\frac{5}{12}$, which corresponds to option (B).