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) : \(\frac{1}{216}\)

Answer 2

We want to find the probability that min(a, b, c) = 6, where a, b, and c are the numbers on the top of three fair dice rolled simultaneously.

This means that all three numbers must be greater than or equal to 6, and at least one of them must be equal to 6.

Since each die has numbers 1 to 6, the only number greater than or equal to 6 is 6 itself. So, all three numbers must be equal to 6.

Thus, we want the probability that a = 6, b = 6, and c = 6.

The probability that a single die shows a 6 is \(\frac{1}{6}\).

Since the three dice are independent, the probability that all three dice show a 6 is the product of the individual probabilities:

\(P(\text{min}(a, b, c) = 6) = P(a = 6, b = 6, c = 6) = P(a = 6)P(b = 6)P(c = 6) = \frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{6} = \frac{1}{216}\)