Question

Two dice are thrown at the same time. Determine the probability that (i) sum of the numbers on the two dice is 5, and (ii) difference of the numbers on the two dice is 3.

• A. (i) \(\frac{1}{9}\), (ii) \(\frac{1}{6}\)
• B. (i) \(\frac{1}{6}\), (ii) \(\frac{1}{9}\)
• C. (i) \(\frac{1}{12}\), (ii) \(\frac{1}{6}\)
• D. (i) \(\frac{1}{9}\), (ii) \(\frac{1}{12}\)

Hint:

When listing outcomes for differences, remember that the order of the dice matters!
If \((1, 4)\) is a favorable outcome, then \((4, 1)\) is also a distinct favorable outcome and must be counted separately.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
Two six-sided dice are thrown simultaneously. We need to determine:
1. The probability that the sum of the numbers showing on the two dice is exactly 5.
2. The probability that the positive difference between the numbers on the two dice is exactly 3.

Step 2: Key Formula or Approach:
1. For two six-sided dice, each die has 6 outcomes. The total number of outcomes in the sample space is:
\[ N = 6 \times 6 = 36 \]
2. Use the classical probability formula for each event:
\[ P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} \]

Step 3: Detailed Explanation:
1.

Part (i): Find the probability that the sum of the numbers is 5:
Let \(E_1\) be the event that the sum of the numbers is 5. We need pairs \((x, y)\) such that \(x + y = 5\), where \(1 \le x, y \le 6\):
- If \(x = 1 \implies y = 4 \implies (1, 4)\)
- If \(x = 2 \implies y = 3 \implies (2, 3)\)
- If \(x = 3 \implies y = 2 \implies (3, 2)\)
- If \(x = 4 \implies y = 1 \implies (4, 1)\)
The number of favorable outcomes is 4.
Calculate the probability:
\[ P(E_1) = \frac{4}{36} = \frac{1}{9} \]

2.

Part (ii): Find the probability that the difference of the numbers is 3:
Let \(E_2\) be the event that the positive difference is 3. We need pairs \((x, y)\) such that \(|x - y| = 3\):
- If \(x = 1 \implies y = 4 \implies (1, 4)\)
- If \(x = 4 \implies y = 1 \implies (4, 1)\)
- If \(x = 2 \implies y = 5 \implies (2, 5)\)
- If \(x = 5 \implies y = 2 \implies (5, 2)\)
- If \(x = 3 \implies y = 6 \implies (3, 6)\)
- If \(x = 6 \implies y = 3 \implies (6, 3)\)
The number of favorable outcomes is 6.
Calculate the probability:
\[ P(E_2) = \frac{6}{36} = \frac{1}{6} \]

Step 4: Final Answer:
The probabilities are \(\frac{1}{9}\) for the sum to be 5, and \(\frac{1}{6}\) for the difference to be 3, which corresponds to option (A).