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Two different dice are thrown together, Find the probability that the numbers obtained have

(i) even sum, and
(ii) even product

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\frac{1}{4}

and

\frac{3}{2}

\frac{1}{2}

and

\frac{3}{4}

\frac{1}{3}

and

\frac{4}{5}

- \frac{1}{2}

and

\frac{2}{3}

answer is B.

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Detailed Solution

Given,
Different dice = 2
When throw the two dice together then
Have numbers = 1, 2, 3, 4, 5, 6 and 1, 2, 3, 4, 5, 6
We have total number of outcomes form the

= 6^{2} = 6^{*} 6 = 36

Outcomes for even sum,

(1, 1), (1, 3), (1, 5)

(2, 2), (2, 4), (2, 6)

(3, 1), (3, 3), (3, 5)

(4, 2), (4, 4), (4, 6)

(5, 1), (5, 3), (5, 5)

(6, 2), (6, 4), (6, 6)

Therefore, favorable outcomes = 18
So, the probability of even sum,

P (A) = \frac{18}{36} = \frac{1}{2}

Hence the probability of even sum =

\frac{1}{2}

.
In the 2nd case we found even product probability. The outcomes in which we get even sum,

(1, 2), (1, 4), (1, 6)

(2, 1), (2, 2), (2, 3) (2, 4), (2, 5), (2, 6)

(3, 2), (3, 4), (3, 6)

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)

(5, 2), (5, 4), (5, 6)

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)

Therefore, favorable outcomes of this case= 27. So the probability of getting even product,

P (B) = \frac{27}{36} = \frac{3}{4}

Hence the probability of even product =

\frac{3}{4}

Correct option is 2.

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