Question

If two dice are thrown simultaneously, then the probability that the sum of the numbers which come up on the dice to be more than 5 is:

• A. \( \frac{5}{36} \)
• B. \( \frac{1}{6} \)
• C. \( \frac{5}{18} \)
• D. \( \frac{7}{18} \)
• E. \( \frac{13}{18} \)

Hint:

For dice sum problems, remember the number of ways to get a sum $S$ increases linearly up to $S=7$ and then decreases. Ways to get sum $S$: $S-1$ for $S \in [2, 7]$.

Answer 1

Answer: (E)

Concept: When two dice are thrown, total outcomes \( = 6 \times 6 = 36 \). It is often easier to find the probability of the complement event (\( \text{sum} \le 5 \)) and subtract it from 1.

Step 1: List outcomes where sum is 5 or less.

• Sum = 2: (1,1) $\rightarrow$ 1 way
• Sum = 3: (1,2), (2,1) $\rightarrow$ 2 ways
• Sum = 4: (1,3), (2,2), (3,1) $\rightarrow$ 3 ways
• Sum = 5: (1,4), (2,3), (3,2), (4,1) $\rightarrow$ 4 ways Total outcomes for sum \( \le 5 = 1 + 2 + 3 + 4 = 10 \).

Step 2: Calculate the probability of the complement.
\[ P(\text{sum} \le 5) = \frac{10}{36} = \frac{5}{18} \]

Step 3: Find the required probability.
\[ P(\text{sum} > 5) = 1 - P(\text{sum} \le 5) \] \[ P(\text{sum} > 5) = 1 - \frac{5}{18} = \frac{13}{18} \]