Question

Two different dice are thrown together. Find the probability that the numbers obtained have
(i) even sum, and
(ii) even product. ​

Answer 1

Answer:

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Step-by-step explanation:

On throwing the two different dice the various outcomes obtained are as follows:

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

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

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

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

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

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

Thus, here total number of outcomes are 36.

(i) Even sum outcomes are as follows:

(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)

Thus, number of favourable outcomes = 18

P(even sum) = 18/36 = 1/2

Hence, the probability of getting an even sum is 1/2.

(ii) Even product outcomes are as follows:

(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)

Thus number of favourable outcomes = 27

P (even product) = 27/36 = 3/4

Hence, the probability of getting an even product is 3/4.

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Answer 2

Answer:

otal no. of outcomes = 6² = 36

1) probability = 18/36 = 3/6=1/2

{ (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) }

2) probability = 27/36 = 9/12=3/4

{ (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) }

Step-by-step explanation: