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Answer: Option 1
In a simultaneous throw of dice, n (S) = (6 × 6) = 36 Let E = event of getting a doublet = [(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)] ∴ P(E) = $$\frac{{n (E)}}{{n (S)}}$$ = $$\frac{{6}}{{36}}$$ = $$\frac{{1}}{{6}}$$
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