Two balanced dice are rolled. Find the probab | vedclass

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(B) When two balanced dice are rolled,the total number of possible outcomes is $6 	imes 6 = 36$.The sum of the numbers on the two dice is odd if one

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$1/2$

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(B) When two balanced dice are rolled,the total number of possible outcomes is $6 	imes 6 = 36$.The sum of the numbers on the two dice is odd if one die shows an even number and the other shows an odd number.Let $E$ be the event that the sum is odd.Possible outcomes for (Odd,Even) are: $(1,2), (1,4), (1,6), (3,2), (3,4), (3,6), (5,2), (5,4), (5,6)$ (Total $9$ outcomes).Possible outcomes for (Even,Odd) are: $(2,1), (2,3), (2,5), (4,1), (4,3), (4,5), (6,1), (6,3), (6,5)$ (Total $9$ outcomes).Total favorable outcomes = $9 + 9 = 18$.Probability $P(E) = \frac{	ext{Favorable outcomes}}{	ext{Total outcomes}} = \frac{18}{36} = \frac{1}{2}$.

Two balanced dice are rolled. Find the probability that the sum of numbers on two dice is odd.

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