If the two spinners shown are each spun once, what is the probability that the sum of the two numbers is either even or a multiple of 3?

Question

Spinner 1 has: (1, 2, 3, 4, 5, 6)

Next spinner has: (1, 2, 3, 4, 5, 6, 7, 8)

Been stuck on this problem for a long time. I know the answer but I have no idea how the textbook got there.

Answer 1

Think of the possible outcomes in spinning each spinner once. I'll label them this way: S1-1 means Spinner #1 got a 1 or S2-8 means Spinner #2 got an 8.

You could have:

S1-1 / S2-1

S1-1 / S2-2

S1-1 / S2-3

S1-1 / S2-4

S1-1 / S2-5

S1-1 / S2-6

S1-1 / S2-7

S1-1 / S2-8

S1-2 / S2-1

S1-2 / S2-2

S1-2 / S2-3

S1-2 / S2-4

S1-2 / S2-5

S1-2 / S2-6

S1-2 / S2-7

S1-2 / S2-8

S1-3 / S2-1

S1-3 / S2-2

S1-3 / S2-3

S1-3 / S2-4

S1-3 / S2-5

S1-3 / S2-6

S1-3 / S2-7

S1-3 / S2-8

S1-4 / S2-1

S1-4 / S2-2

S1-4 / S2-3

S1-4 / S2-4

S1-4 / S2-5

S1-4 / S2-6

S1-4 / S2-7

S1-4 / S2-8

S1-5 / S2-1

S1-5 / S2-2

S1-5 / S2-3

S1-5 / S2-4

S1-5 / S2-5

S1-5 / S2-6

S1-5 / S2-7

S1-5 / S2-8

S1-6 / S2-1

S1-6 / S2-2

S1-6 / S2-3

S1-6 / S2-4

S1-6 / S2-5

S1-6 / S2-6

S1-6 / S2-7

S1-6 / S2-8

That is all the possible choices. There are 48 of them 6 x 8 = 48

You can count up the choices where the sum is even (both numbers must be even or both numbers must be odd for this to happen). In the S1-1 group there are 4. In the S1-2 group there are also 4, keep doing this and you find that half of the choices result in even sums. So there are 24 of these.

Since we have already counted all these, we can ignore them when we are looking for multiples of 3. Here are the ones that are left and I'll put their sums the the right of each:

S1-1 / S2-2 3

S1-1 / S2-4 5

S1-1 / S2-6 7

S1-1 / S2-8 9

S1-2 / S2-1 3

S1-2 / S2-3 5

S1-2 / S2-5 7

S1-2 / S2-7 9

S1-3 / S2-2 5

S1-3 / S2-4 7

S1-3 / S2-6 9

S1-3 / S2-8 11

S1-4 / S2-1 5

S1-4 / S2-3 7

S1-4 / S2-5 9

S1-4 / S2-7 11

S1-5 / S2-2 7

S1-5 / S2-4 9

S1-5 / S2-6 11

S1-5 / S2-8 13

S1-6 / S2-1 7

S1-6 / S2-3 9

S1-6 / S2-5 11

S1-6 / S2-7 13

Notice that the only ones that are multiples of 3 are 3 and 9 and there are 8 of those.

8 plus the previous 24 = 32

So the probability is 32/48 or 2/3

Answer 2

There are 6 * 8 = 48 possible pairs in the sample space of results.

(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (1,7), (1,8)

(2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (2,7), (2,8)

(3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (3,7), (3,8)

(4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (4,7), (4,8)

(5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (5,7), (5,8)

(6,1), (6,2), (6,3), (6,4), (6,5), (6,6), (6,7), (6,8)

sums:

2, 3, 4, 5, 6, 7, 8, 9

3, 4, 5, 6, 7, 8, 9,10

4, 5, 6, 7, 8, 9, 10,11

5, 6, 7, 8, 9, 10, 11, 12

6, 7, 8, 9, 10, 11, 12, 13

7, 8, 9, 10, 11, 12, 13, 14

How many of those are even? 24 (logical since there are 48 possible sums)

How many are multiples of 3? (sum would have to be 3, 6, 9 or 12): 16 (logical since that is 1/3 of 48).

How many are one or the other: 16 + 24 = 40

probability of being one or the other is 40/48 = 5/6