Question

Two dice are thrown together. The probability of getting a sum more than 8 is

• A. \( \frac{7}{36} \)
• B. \( \frac{5}{12} \)
• C. \( \frac{5}{18} \)
• D. \( \frac{7}{18} \)

Hint:

To find the probability of a specific event, count the favorable outcomes and divide by the total number of possible outcomes.

Answer 1

The Correct Option is C

Step 1: Total outcomes when two dice are thrown: \( 6\times6=36 \).
Step 2: Count outcomes with sum 8 or less instead: sums 2 through 8 give \( 1+2+3+4+5+6+5=26 \) outcomes.
Step 3: Favorable outcomes for sum \(>8\) are the remaining \( 36-26=10 \), so the probability is \( \frac{10}{36}=\frac{5}{18} \).
\[ \boxed{P(\text{sum}>8) = \frac{5}{18}} \]

Answer 2

A third way to confirm this is to lay out every one of the 36 equally likely ordered pairs \( (d_1,d_2) \) and directly mark which ones have a sum greater than 8, rather than using a shortcut formula.

Listing pairs with sum greater than 8: for sum 9, \( (3,6),(4,5),(5,4),(6,3) \); for sum 10, \( (4,6),(5,5),(6,4) \); for sum 11, \( (5,6),(6,5) \); for sum 12, \( (6,6) \). Counting these explicitly listed pairs: \( 4+3+2+1=10 \) pairs out of the full 36 equally likely pairs.

1. \( \frac{7}{36} \): Would require only 7 such pairs, but the explicit list above shows 10.
2. \( \frac{5}{12} \): Equals \( \frac{15}{36} \), which would require 15 listed pairs, five more than actually exist.
3. \( \frac{5}{18} \): Equals \( \frac{10}{36} \), matching the 10 pairs listed exactly above.
4. \( \frac{7}{18} \): Equals \( \frac{14}{36} \), again more pairs than the list actually contains.

The direct enumeration of all favorable pairs confirms there are exactly 10 out of 36 outcomes with a sum greater than 8.

Therefore, the correct answer is \( \frac{5}{18} \).