MHT CET · Maths · Probability
Two dice are rolled, If both dice have six faces numbered \(1,2,3,5,7,11\), then the probability that the sum of the numbers on upper most face is prime is
- A \(\frac{1}{4}\)
- B \(\frac{3}{4}\)
- C \(\frac{1}{9}\)
- D \(\frac{2}{7}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{4}\)
Step-by-step Solution
Detailed explanation
Two dice are rolled.
\(\therefore \quad \mathrm{n}(\mathrm{S})=36\)
A : Event that the sum of the numbers on upper most face is prime.
\(\begin{aligned} \therefore \quad & A=\{(1,1),(1,2),(2,1),(2,3),(2,5),(2,11), \\ & (3,2),(5,2),(11,2)\}\end{aligned}\)
\(\therefore \quad \mathrm{P}(\mathrm{A})=\frac{9}{36}=\frac{1}{4}\)
\(\therefore \quad \mathrm{n}(\mathrm{S})=36\)
A : Event that the sum of the numbers on upper most face is prime.
\(\begin{aligned} \therefore \quad & A=\{(1,1),(1,2),(2,1),(2,3),(2,5),(2,11), \\ & (3,2),(5,2),(11,2)\}\end{aligned}\)
\(\therefore \quad \mathrm{P}(\mathrm{A})=\frac{9}{36}=\frac{1}{4}\)
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