Question

A pair of fair dice is rolled. What is the probability that the second die lands on a higher value than the first?

• A. \(\frac{1}{36}\)
• B. \(\frac{5}{36}\)
• C. \(\frac{1}{6}\)
• D. \(\frac{5}{12}\)
• E.

Hint:

By symmetry, \(P(y > x) = P(x > y)\). Since there are 6 outcomes where \(x = y\), the number of outcomes where \(x \neq y\) is \(36 - 6 = 30\). Thus, \(P(y > x) = \frac{30/2}{36} = \frac{15}{36} = \frac{5}{12}\).

Answer 1

The Correct Option is D

Concept: When two fair dice are rolled, the total number of possible outcomes is: \[ 6 \times 6 = 36 \] Each outcome can be written as an ordered pair \((x,y)\), where \(x\) = value on the first die, \(y\) = value on the second die. We need to find the probability that \(y > x\).

Step 1: List favorable outcomes. We count the pairs \((x,y)\) where the second value is greater than the first:
• If \(x=1\), \(y \in \{2,3,4,5,6\}\) (5 outcomes)
• If \(x=2\), \(y \in \{3,4,5,6\}\) (4 outcomes)
• If \(x=3\), \(y \in \{4,5,6\}\) (3 outcomes)
• If \(x=4\), \(y \in \{5,6\}\) (2 outcomes)
• If \(x=5\), \(y \in \{6\}\) (1 outcome) Total favorable outcomes: \[ 5 + 4 + 3 + 2 + 1 = 15 \]

Step 2: Compute the probability. \[ P = \frac{\text{Favourable outcomes}}{\text{Total outcomes}} = \frac{15}{36} \] Simplifying the fraction by dividing by 3: \[ P = \frac{5}{12} \]

Step 3: State the final answer. The probability is: \[ \boxed{\frac{5}{12}} \]