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(C) When a balanced die is rolled twice,the total number of possible outcomes is $6 	imes 6 = 36$.The even numbers on a die are $\{2, 4, 6\}$.The eve

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$\frac{1}{4}$

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(C) When a balanced die is rolled twice,the total number of possible outcomes is $6 	imes 6 = 36$.The even numbers on a die are $\{2, 4, 6\}$.The event of getting an even number both times means the outcome must be an ordered pair $(x, y)$ where both $x$ and $y$ are from the set $\{2, 4, 6\}$.The favourable outcomes are: $(2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6)$.The number of favourable outcomes is $m = 9$.Therefore,the required probability $P = \frac{	ext{Number of favourable outcomes}}{	ext{Total number of outcomes}} = \frac{9}{36} = \frac{1}{4}$.

$A$ balanced die is rolled twice. Then,the probability of the event that an even number is received both the times is ..........

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