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A die is thrown thrice. If the first throw is a four, then the chance of getting 11 as total sum, is

Solution

✅ Correct Option: 3

Given the first die shows 4, the remaining two throws must total 7. Ordered pairs summing to 7 number 6 out of 36 equally likely pairs, so the required probability is 636=16\frac{6}{36}=\frac{1}{6}.

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