Question

Three fair dice are rolled simultaneously. Let a, b, c be the numbers on the top of the dice. Then the probability that min(a, b, c) =6 is

• A. $\frac{1}{216}$
• B. $\frac{1}{36}$
• C. $\frac{1}{6}$
• D. $\frac{11}{216}$
• E. $\frac{5}{6}$

Answer 1

The Correct Option is A

We are given that three fair dice are rolled simultaneously. Let $a, b, c$ be the numbers on the top of the dice. We are asked to find the probability that the minimum value among $a, b, c$ is 6.
Step 1: Understanding the condition For $\min(a, b, c) = 6$, this means that all three dice must show at least 6, and the smallest value among $a, b, c$ should be exactly 6. In other words, at least one die must show a 6, and no die should show a value less than 6. Therefore, the only possible outcome for $a, b, c$ is that one die shows a 6 and the other two dice must show 6 as well.
Step 2: Count the number of favorable outcomes For $\min(a, b, c) = 6$, all three dice must show 6. There is only one favorable outcome: $(6, 6, 6)$.
Step 3: Count the total number of outcomes Since each die has 6 faces, the total number of possible outcomes when rolling 3 dice is: $$6 \times 6 \times 6 = 216.$$
Step 4: Calculate the probability The probability is the ratio of favorable outcomes to the total number of possible outcomes.
Since there is only 1 favorable outcome ($(6, 6, 6)$) out of 216 possible outcomes, the probability is: $$\frac{1}{216}.$$

The correct option is (A) : $\frac{1}{216}$