Question:
Two dies are rolled. If both dices have six faces numbered 1, 2, 3, 5, 7 and 11, then the probability that the sum of the numbers on the top faces is less than or equal to 8 is :
Two dies are rolled. If both dices have six faces numbered 1, 2, 3, 5, 7 and 11, then the probability that the sum of the numbers on the top faces is less than or equal to 8 is :
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There are 36 equally likely ordered pairs; count those with sum at most 8.
Updated On: Nov 4, 2026
- \(\frac{4}{9}\)
- \(\frac{1}{2}\)
- \(\frac{5}{12}\)
- \(\frac{17}{36}\)
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The Correct Option is D
Solution and Explanation
Total outcomes \(= 6 \times 6 = 36\). Count ordered pairs \((a, b)\) from \(\{1, 2, 3, 5, 7, 11\}\) with \(a + b \le 8\):
\(a = 1\): \(b \in \{1, 2, 3, 5, 7\}\) gives 5
\(a = 2\): \(b \in \{1, 2, 3, 5\}\) gives 4
\(a = 3\): \(b \in \{1, 2, 3, 5\}\) gives 4
\(a = 5\): \(b \in \{1, 2, 3\}\) gives 3
\(a = 7\): \(b \in \{1\}\) gives 1
\(a = 11\): none, 0
Favourable \(= 5 + 4 + 4 + 3 + 1 = 17\).
Probability \(= \frac{17}{36}\).
Answer: (4) \(\frac{17}{36}\)
\(a = 1\): \(b \in \{1, 2, 3, 5, 7\}\) gives 5
\(a = 2\): \(b \in \{1, 2, 3, 5\}\) gives 4
\(a = 3\): \(b \in \{1, 2, 3, 5\}\) gives 4
\(a = 5\): \(b \in \{1, 2, 3\}\) gives 3
\(a = 7\): \(b \in \{1\}\) gives 1
\(a = 11\): none, 0
Favourable \(= 5 + 4 + 4 + 3 + 1 = 17\).
Probability \(= \frac{17}{36}\).
Answer: (4) \(\frac{17}{36}\)
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