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Question
In a simultaneous throw of a pair of dice, find the probability of getting a doublet of prime numbers
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Solution
We know that in a single throw of two dices, the total number of possible outcomes is (6 × 6) = 36.
Let S be the sample space.
Then n(S) = 36
Let E3 = event of getting a doublet of prime numbers
Then E3 = {(2, 2), (3, 3), (5, 5)}
i.e. n (E3) = 3
\[\therefore P\left( E_3 \right) = \frac{n\left( E_3 \right)}{n\left( S \right)} = \frac{3}{36} = \frac{1}{12}\]
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