If two balanced dice are tossed once,the prob | vedclass

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(C) When two balanced dice are tossed,the total number of possible outcomes is $6 	imes 6 = 36$.Let $E$ be the event that the sum of the integers on

Vedclass · Accepted Answer

$\frac{1}{9}$

Vedclass · Answer

(C) When two balanced dice are tossed,the total number of possible outcomes is $6 	imes 6 = 36$.Let $E$ be the event that the sum of the integers on the upper faces is $9$.The possible outcomes for the sum to be $9$ are $(3, 6), (4, 5), (5, 4), 	ext{ and } (6, 3)$.Thus,the number of favorable outcomes is $4$.The probability of the event $E$ is given by $P(E) = \frac{	ext{Number of favorable outcomes}}{	ext{Total number of outcomes}} = \frac{4}{36} = \frac{1}{9}$.

If two balanced dice are tossed once,the probability of the event that the sum of the integers appearing on the upper faces of the two dice is $9$ is:

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