In a single throw of two dice,the probability | vedclass

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(B) When two dice are thrown,the total number of possible outcomes is $6 	imes 6 = 36$.An odd sum is obtained if one die shows an odd number and the

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$\frac{1}{2}$

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(B) When two dice are thrown,the total number of possible outcomes is $6 	imes 6 = 36$.An odd sum is obtained if one die shows an odd number and the other shows an even number,or vice versa.The possible outcomes for an odd sum are: $(1,2), (1,4), (1,6), (2,1), (2,3), (2,5), (3,2), (3,4), (3,6), (4,1), (4,3), (4,5), (5,2), (5,4), (5,6), (6,1), (6,3), (6,5)$.Counting these,we find there are $18$ such outcomes.The probability $P$ of obtaining an odd sum is $\frac{	ext{Number of favorable outcomes}}{	ext{Total number of outcomes}} = \frac{18}{36} = \frac{1}{2}$.

In a single throw of two dice,the probability of obtaining an odd number as the sum is

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