Question

A die is thrown twice. The probability of getting a sum equal to 8 is:

• A. \( \frac{1}{12} \)
• B. \( \frac{5}{36} \)
• C. \( \frac{1}{6} \)
• D. \( \frac{7}{36} \)

Hint:

For problems involving the sum of two dice, the number of ways to get a sum \(S\) (where \(2 \le S \le 7\)) is always \((S - 1)\). For a sum \(S\) (where \(8 \le S \le 12\)), the number of ways is \((13 - S)\). Here, for a sum of 8, the number of ways is quickly found as \(13 - 8 = 5\).

Answer 1

The Correct Option is B

Concept: When a single fair six-sided die is thrown twice, the sample space consists of ordered pairs \((i, j)\), where \(i\) represents the outcome of the first throw and \(j\) represents the outcome of the second throw. The total number of possible outcomes in the sample space is: \[ n(S) = 6 \times 6 = 36 \] The probability of an event \(E\) is given by the ratio of the number of favorable outcomes \(n(E)\) to the total number of outcomes \(n(S)\): \[ P(E) = \frac{n(E)}{n(S)} \]

Step 1: Identifying the favorable outcomes for a sum of 8.
Let \(E\) be the event of getting a sum equal to 8. We look for pairs \((i, j)\) such that \(i + j = 8\), where \(1 \le i, j \le 6\). Tracing out the possible outcomes systematically:

• If the first die shows \(2\), the second die must show \(6 \implies (2, 6)\)
• If the first die shows \(3\), the second die must show \(5 \implies (3, 5)\)
• If the first die shows \(4\), the second die must show \(4 \implies (4, 4)\)
• If the first die shows \(5\), the second die must show \(3 \implies (5, 3)\)
• If the first die shows \(6\), the second die must show \(2 \implies (6, 2)\)

Note that pairs like \((1, 7)\) are impossible because a standard die only has numbers up to 6. Thus, the set of favorable outcomes is: \[ E = \{(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)\} \] The total number of favorable outcomes is \(n(E) = 5\).

Step 2: Calculating the probability.
Substitute the values into the probability formula: \[ P(E) = \frac{n(E)}{n(S)} = \frac{5}{36} \]