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:
We are considering an event where three standard fair dice are rolled at once.
The variables \( a, b, \) and \( c \) denote the discrete outcomes of each individual die, meaning they can only take integer values from 1 to 6.
The mathematical function \( \min(a, b, c) \) evaluates to the smallest value among the three numbers.
We need to compute the probability of the specific condition where this smallest value is exactly equal to 6.
Step 2: Key Formula or Approach:
1. Total Outcomes: For rolling \( n \) dice, the total number of elementary events in the sample space is \( 6^n \).
2. Inequality Logic for Min Function: The condition \( \min(x_1, x_2, \dots, x_n) \ge k \) mathematically requires that every single variable satisfies the condition \( x_i \ge k \).
3. Probability: \( P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} \).
Step 3: Detailed Explanation:
Let's first determine the total number of possible outcomes when rolling three dice.
Since each die operates independently and has 6 faces, the fundamental counting principle applies:
\[ \text{Total Outcomes} = 6 \times 6 \times 6 = 216 \] Now, let's analyze the favorable condition: \( \min(a, b, c) = 6 \).
By definition, if the minimum of a set of numbers is 6, it means that absolutely no number in that set can be smaller than 6.
Mathematically, this forces the following simultaneous inequalities:
\[ a \ge 6 \] \[ b \ge 6 \] \[ c \ge 6 \] However, we know that \( a, b, \) and \( c \) are the results of standard dice rolls, so their maximum possible value is constrained to 6.
Therefore, the inequalities strictly collapse into equalities:
\[ a = 6, \quad b = 6, \quad c = 6 \] This signifies that all three dice must independently land on the number 6.
There is only one specific outcome that satisfies this requirement, which is the triplet \( (6, 6, 6) \).
\[ \text{Number of favorable outcomes} = 1 \] Finally, compute the probability by dividing the favorable outcomes by the total outcomes:
\[ P = \frac{1}{216} \] Step 4: Final Answer:
The probability is \( \frac{1}{216} \).