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Two dice are rolled and given that both faces are showing even numbers. The probability that their sum is more than 9 is

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$\frac{1}{3}$

$\frac{2}{9}$

$\frac{1}{2}$

$\frac{1}{4}$

answer is A.

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Detailed Solution

$n (E) = {(4, 6) (6, 4) (6, 6)} = 3$

n(s)={(2,2) (2,4) (2,6) (4,2) (4,4) (4,6) (6,2) (6,4) (6,6)}=9

$P (E) = \frac{3}{9} = \frac{1}{3}$

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