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. \(\frac{5}{6}\)

Hint:

The probability that \(\min(X_1, \dots, X_n) \ge k\) is \(P(X \ge k)^n\). Here, \(P(X \ge 6) = 1/6\). So, \((1/6)^3 = 1/216\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The notation \(\min(a, b, c) = 6\) means that the smallest value among the three dice must be 6. Since the maximum value on a standard die is 6, the only way for the minimum to be 6 is if every single die shows a 6.

Step 2: Key Formula or Approach
1. Total outcomes for three dice = \(6^3 = 216\).
2. Condition: \(a \ge 6, b \ge 6, c \ge 6\).

Step 3: Detailed Explanation
1. On a die, the possible values are \(\{1, 2, 3, 4, 5, 6\}\).
2. If the minimum value is 6, then: - \(a\) must be 6. - \(b\) must be 6. - \(c\) must be 6.
3. There is only \(1 \times 1 \times 1 = 1\) favorable outcome: \((6, 6, 6)\).
4. Probability = \(\frac{\text{Favorable}}{\text{Total}} = \frac{1}{216}\).

Step 4: Final Answer
The probability is \(\frac{1}{216}\).