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 $6 	imes 6 = 36$.We are looking for the event where the sum of the num

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$\frac{1}{9}$

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(C) When two balanced dice are tossed,the total number of possible outcomes is $6 	imes 6 = 36$.We are looking for the event where the sum of the numbers on the upper faces is $9$.The possible pairs $(d_1, d_2)$ that result in a sum of $9$ are: $(3, 6), (4, 5), (5, 4), (6, 3)$.There are $4$ such favorable outcomes.The probability of the event is given by the ratio of favorable outcomes to total outcomes:$P = \frac{	ext{Number of favorable outcomes}}{	ext{Total number of outcomes}} = \frac{4}{36} = \frac{1}{9}$.

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

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