MHT CET · Maths · Probability
Two dice are rolled simultaneously. The probability that the sum of the two numbers on the dice is a prime number, is
- A \(\frac{5}{11}\)
- B \(\frac{5}{12}\)
- C \(\frac{7}{12}\)
- D \(\frac{7}{11}\)
Answer & Solution
Correct Answer
(B) \(\frac{5}{12}\)
Step-by-step Solution
Detailed explanation
The sum of numbers on the two dice can be \((2,3, \ldots ., 12)\) and prime numbers in this list are 2, 3, 5, 7, 11 .
\(2 \Rightarrow(1,1)\) and \(3 \Rightarrow(1,2),(2,1)\) and 5
\(\Rightarrow(1,4),(2,3),(3,2),(4,1)\),
\(7 \Rightarrow(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\) and 11
\(\Rightarrow(5,6),(6,5)\)
Thus \(\mathrm{n}(\mathrm{s})=6 \times 6=36\) and \(\mathrm{n}(\mathrm{E})=15\)
Hence required probability \(=\frac{15}{36}=\frac{5}{12}\)
\(2 \Rightarrow(1,1)\) and \(3 \Rightarrow(1,2),(2,1)\) and 5
\(\Rightarrow(1,4),(2,3),(3,2),(4,1)\),
\(7 \Rightarrow(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\) and 11
\(\Rightarrow(5,6),(6,5)\)
Thus \(\mathrm{n}(\mathrm{s})=6 \times 6=36\) and \(\mathrm{n}(\mathrm{E})=15\)
Hence required probability \(=\frac{15}{36}=\frac{5}{12}\)
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