Question

Wer +266. A bag contains $3$ balls labeled $2, 4$ and $8$. A ball is to be picked, the value on the label is to be recorded and then the ball is to be returned to the bag. This will be done three times and then the values will be added together. What is the sum of all of the possible distinct sums?

Answer 1

Answer: 128

Step-by-step explanation:

Any sum formed by a combination of the numbers 2, 4, 8 and must be divisible by 2. The smallest possible value of such a sum is equal to 3*2=6 , and the largest possible value of such a sum is equal to 3*8=24 . After testing, we find that

6 = 2+2+2, 8=4+2+2, 10=4+4+2

12 = 4+4+4, 14 = 8+4+2, 16 = 8+4+4

18 = 8+8+2, 20 = 8+8+4, 24 = 8+8+8

However, we cannot find a combination that will add to be 22: if two of the numbers are not 8, then the maximum possible sum is 4+4+8 = 16. Thus, two of the numbers picked must be 8, but then the third ball must have the number 6, which is not possible. Thus, the answer is the sum of the even numbers from 6 to 24 excluding 22, which is 128 .

Answer 2

The sum of all the possible distinct sum is 128

Explanation:

The number of balls in the bag = 3

The ball labels, (the numbers written on the balls) = 2, 4, and 8

The number of balls selected with replacement = 1

Therefore, we have have, the number of ways of selecting 1 ball from 3 = 3 ways

The distinct combination of the selected balls are;

2, 2, 2 with sum 2 + 2 + 2 = 6

2, 2, 4 with sum 2 + 2 + 4 = 8

2, 2, 8 with sum 2 + 2 + 8 = 12

2, 4, 4 with sum 2 + 4 + 4 = 10

2, 4, 8 with sum 2 + 4 + 8 = 14

2, 8, 8 with sum 2 + 8 + 8 = 18

4, 4, 4 with sum 4 + 4 + 4 = 12

4, 4, 8 with sum 4 + 4 + 8 = 16

8, 8, 4 with sum 8 + 8 + 4 = 20

8, 8. 8 with sum 8 + 8 + 8 = 24

The distinct sums are therefore;

6, 8, 12, 10, 14, 18, 16, 20, 24

The sum of the distinct sum is 6 + 8 + 12 + 10 + 14 + 18 + 16 + 20 + 24 = 128.