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Six faces of a die are marked with the number $1, - 1, 0, - 2, 2, 3$ and the die is thrown thrice . The probability that the sum of the numbers thrown is six is

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$\frac{3}{216}$

$\frac{6}{216}$

$\frac{10}{216}$

$\frac{18}{216}$

answer is C.

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Detailed Solution

$1 + 2 + 3 \to can be interchanged in 3! = 6 wavs 0 + 3 + 3 \to can be interchanged in 3! / 2! = 3 ways 2 + 2 + 2 \to= 1 way$

$n (s) = 216, n (E) = 1 + 6 + 3 = 10$

$p (E) = \frac{10}{216}$

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