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A dice is thrown twice. Find the probability of getting an odd number in the second throw and a multiple of 3 in the first throw.

Solution

Correct Option: 3

P(multiple of 3 in first throw) = 2/6 = 1/3 (outcomes 3, 6).

P(odd in second throw) = 3/6 = 1/2 (outcomes 1, 3, 5).

Since throws are independent, combined probability = 13×12=16\frac{1}{3} \times \frac{1}{2} = \frac{1}{6}.

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