Question

A hat contains 5 balls. The balls are numbered 1, 2, 5, 7, and 10. One ball is randomly selected and not replaced, and then a second ball is selected. The numbers on the 2 balls are added together.

A fair decision is to be made about which one of two candy bars to purchase, using the sum of the numbers on the balls.
The candy bar options are Choco Delight or Go Nuts.
Which description accurately explains how a fair decision can be made in this situation?

If the sum of the balls is less than or equal to 9, purchase Choco Delight. If the sum is greater than 9, purchase Go Nuts.
If the sum of the balls is a factor of 40, purchase Choco Delight. If the sum is not a factor of 40, purchase Go Nuts.
If the sum is a multiple of 3, purchase Choco Delight. If the sum is not a multiple of 3, purchase Go Nuts.
If the sum of the balls is even, purchase Choco Delight. If the sum is odd, purchase Go Nuts.

Answer 1

Answer: Choice A

If the sum of the balls is less than or equal to 9, purchase Choco Delight. If the sum is greater than 9, purchase Go Nuts.

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Step-by-step explanation:

Use a spreadsheet or graph paper, to make a table with 5 rows and 5 columns. Label the headers as 1,2,5,7,10

Cross out the northwest main diagonal and everything below it. The cells not crossed out represent the different combos of sums possible.

Here's all of those combos:

• 1+2 = 3
• 1+5 = 6
• 1+7 = 8
• 1+10 = 11
• 2+5 = 7
• 2+7 = 9
• 2+10 = 12
• 5+7 = 12
• 5+10 = 15
• 7+10 = 17

The sums, listed from smallest to largest, are: {3,6,7,8,9,11,12,12,15,17}

I kept the duplicate in there to show there are 10 sums we can reach.

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Let's break things into two groups. We'll have stuff that's 9 or smaller in one group, and then stuff larger than 9 in another group.

• less than or equal to nine: 3,6,7,8,9
• greater than nine: 11,12,12,15,17

Each group has the same number of values (5), which means this is a fair way to determine what kind of candy bar to pick. Each group is equally likely to be chosen. It's equivalent to a coin flip. This is why choice A is the final answer.

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Choices B through D lead to some imbalance which means it won't be a fair process. They can be ruled out.

For example, choice C will have these two groups

• multiples of three: 3,6,9,12,12,15
• non-multiples of three: 7,8,11,17

The first group has 6 values and the second group has 4 values. I'll let you check the others.