Question

Three fair dice are thrown simultaneously. What is the probability that the sum is 5?

• A. 1/36
• B. 1/6
• C. 1/54
• D. 1/72

Hint:

For probability problems with dice, first calculate the total possible outcomes, then count the favorable outcomes to find the probability.

Answer 1

The Correct Option is A

Listing every way three dice can sum to 5 gives just two number-sets, (1,1,3) and (1,2,2).
Each set can land on the three dice in 3 different orders, so there are \( 3+3=6 \) favorable outcomes.
The total number of outcomes for three dice is \( 6^3=216 \).
\[ P = \frac{6}{216} = \frac{1}{36} \]
Therefore, the correct answer is 1/36.

Answer 2

A more formula-driven way to count the favorable outcomes is to use the standard result for the number of ways three dice (each showing 1 to 6) can sum to a target value, then convert that count to a probability.

1. 1/36: Substituting each die's value as \( 1 \le x_i \le 6 \) and setting \( x_1+x_2+x_3=5 \), a change of variable \( y_i = x_i - 1 \) turns this into \( y_1+y_2+y_3=2 \) with each \( y_i \ge 0 \), and the number of non-negative integer solutions to this equation is \( \binom{2+2}{2}=6 \), with no need to check the upper bound of 6 since the target sum is so low. This gives exactly 6 favorable outcomes out of 216, or \( \frac{1}{36} \).
2. 1/6: This would need 36 solutions to the equation above, far more than the 6 that the counting formula actually gives.
3. 1/54: This corresponds to only 4 solutions, fewer than the correctly counted 6.
4. 1/72: This corresponds to only 3 solutions, again fewer than the correctly counted 6.

The combinatorial formula confirms 6 favorable outcomes out of 216 total.

Therefore, the correct answer is 1/36.