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

\frac{3}{216}

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\frac{6}{216}

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\frac{10}{216}

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\frac{18}{216}

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Solution

The correct option is D $\frac{10}{216}$
Six faces are marked as 1, -1, 0, -2, 2, 3
Die is thrown thrice.
The following are possibilities for the sum to be 6

$1 + 2 + 3 \to$ can be interchanged in 3!=6 ways
$0 + 3 + 3 \to$ can be interchanged in 3!/2!=3 ways
$2 + 2 + 2 \to$ = 1 way
So,number of ways = 10 ways
Total number of possibilities = 666
= 216

Probability = $\frac{10}{216}$

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