A balanced die is rolled twice. Then,the prob | vedclass

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(A) When a balanced die is rolled twice,the total number of possible outcomes is $6 	imes 6 = 36$.Let $E$ be the event that the sum of the numbers ob

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$\frac{1}{12}$

Vedclass · Answer

(A) When a balanced die is rolled twice,the total number of possible outcomes is $6 	imes 6 = 36$.Let $E$ be the event that the sum of the numbers obtained in two trials is more than $10$.The outcomes where the sum is greater than $10$ are $(5, 6)$,$(6, 5)$,and $(6, 6)$.Thus,the number of favorable outcomes is $m = 3$.The probability $P(E)$ is given by the ratio of favorable outcomes to total outcomes:$P(E) = \frac{m}{n} = \frac{3}{36} = \frac{1}{12}$.

$A$ balanced die is rolled twice. Then,the probability of the event that the sum of the numbers received in two trials is more than $10$ is ...........

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