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(D) When a balanced die is rolled twice,the total number of possible outcomes is $6 	imes 6 = 36$.The sum of the numbers obtained in two trials is ev

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$\frac{1}{2}$

Vedclass · Answer

(D) When a balanced die is rolled twice,the total number of possible outcomes is $6 	imes 6 = 36$.The sum of the numbers obtained in two trials is even if:$1$. Both numbers are even (e.g.,$2, 4, 6$ for each die: $3 	imes 3 = 9$ outcomes).$2$. Both numbers are odd (e.g.,$1, 3, 5$ for each die: $3 	imes 3 = 9$ outcomes).Total number of favorable outcomes $= 9 + 9 = 18$.Therefore,the required probability $= \frac{	ext{Number of favorable outcomes}}{	ext{Total number of outcomes}} = \frac{18}{36} = \frac{1}{2}$.

$A$ balanced die is rolled twice. Then,the probability of the sum of numbers received in these two trials being even is ...........

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