Question

Stacy rolls a pair of six-sided fair dice. The probability that the sum of the numbers rolled is either a multiple of 3 or an even number is , and the two events are exclusive

Answer 1

3,6 are the only multiples so it would be 2/6 but simplify that to 1/3
2,4,6 are evens so it would be 3/6 which is also 1/2

Answer 2

Answer:

5/9

Step-by-step explanation:

Let the probability of getting a multiple of 3 is "A"

Let the probability of getting an even number is "B"

Total number of outcomes=36

Total number of favourable outcomes  of either a multiple of 3 or an even number are

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

That is we have 20 favourable outcomes

Hence, $Probability=\frac{\text{favourable outcomes}}{\text{total number of outcomes}}$

Therefore, after substituting the values we get

$probability=\frac{20}{36}$

After simplification we will get

$probability=\frac{5}{9}$