Three fair dice are rolled simultaneously. Find the probability that the sum of the numbers shown on the dice is exactly $10$ .

medium 2 marks Class 12 · Mathematics · Probability

Answer

\frac{1}{8}

The correct answer is $\frac{1}{8}$ — Step 6: Probability = $\frac{27}{216} = \frac{1}{8}$ .

Step-by-step solution

Step 1: Total possible outcomes = $6^{3} = 216$ .
Step 2: Convert the condition $a + b + c = 10$ into non‑negative variables by setting $a^{'} = a - 1$ , $b^{'} = b - 1$ , $c^{'} = c - 1$ . Then $a^{'} + b^{'} + c^{'} = 7$ with $0 l e a^{'}, b^{'}, c^{'} l e 5$ .
Step 3: Count all non‑negative solutions to $a^{'} + b^{'} + c^{'} = 7$ using stars and bars: $(3 - 1 7 + 3 - 1) = (2 9) = 36$ .
Step 4: Subtract solutions where a variable exceeds $5$ . For a variable $> 5$ , set it to $6 + x$ ; the remaining sum is $1$ , giving $(3 - 1 1 + 3 - 1) = (2 3) = 3$ solutions. There are 3 such variables, so subtract $3 \times 3 = 9$ .
Step 5: No double over‑counting is possible because two variables cannot both exceed $5$ simultaneously. Thus favorable outcomes = $36 - 9 = 27$ .
Step 6: Probability = $\frac{27}{216} = \frac{1}{8}$ .

Common mistake: Counting only combinations and ignoring order, or overlooking the restriction each die shows at most 6

Concept tested: Discrete uniform distribution – dice sums

Published 11 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked