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Thanks to James W. for the suggestion!
At an amusement park, 65% of visitors ate a donut, 80% ate a soft pretzel, 80% ate pizza, and 90% ate ice cream. What is the minimum percentage of visitors that ate all 4 foods?
As usual, watch the video for a solution.
Logic Test – How Many Ate All Foods?
Or keep reading.
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“All will be well if you use your mind for your decisions, and mind only your decisions.” Since 2007, I have devoted my life to sharing the joy of game theory and mathematics. MindYourDecisions now has over 1,000 free articles with no ads thanks to community support! Help out and get early access to posts with a pledge on Patreon.
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Answer To Minimum Percentage of Four Groups Question
(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).
Method 1
Imagine there are 20 visitors to the park, so 13 ate a donut, 16 ate a pretzel, 16 ate pizza, and 18 ate ice cream. We can minimize the overlap of all 4 foods by forcing as many people to eat 3 foods but not the 4th. So suppose the 7 people who did not have a donut ate all other foods, the 4 people who ate a pretzel ate the other three, etc. Visually it looks like this:
Still we must have that the first 3 people did eat all 4 foods. So that is 3/20 = 15%.
Method 2
We have these percentages
65% donut
80% pretzel
80% pizza
90% ice cream
Consider the minimum number that ate both donuts and pretzels. There are 35% of people who did not have a donut. If all of them had a pretzel, then we still have 80 – 35 = 45% who ate a pretzel and a donut.
Similarly, there are 20% of people who did not have pizza. If all of them had ice cream, then still there are 90 – 20 = 70% of people who had pizza and ice cream.
So imagine these results are two new foods:
45% donut-pretzel
70% pizza-ice cream
Make all of the 55% who did not have donut-pretzel have pizza-ice cream. That still leaves 70 – 55 = 15% of people who had both. This is the minimum percentage of people who had all 4 foods.
Method 3
Imagine there are 100 people so the percentages are the number of each food eaten. Totaling the values we get 65 + 80 + 80 + 90 = 315 food items. If all 100 people ate exactly 3 items, then 300 food items would be eaten. We are 15 short. Each extra item has to be eaten by an additional person since each person can eat 4 foods at most. So we must have at least 15 people who ate all 4 food items, which is 15/100 = 15% of visitors.
Minimality
Imagine we have 100 people and sets for how many items eaten:
65 donut W
80 pretzel X
80 pizza Y
90 ice cream Z
Consider all sets to be subsets of {1, 2, …, 100}. Let Ac denote the complement of set A. With 100 people, we we have:
35 donut Wc
20 pretzel Xc
20 pizza Yc
10 ice cream Zc
For any sets we have basic identities:
|A ∪ B| ≤ |A| + |B|
|A ∩ B| = 100 – |Ac ∪ Bc|
We wish to find the intersection of all four sets, so we have:
|W ∩ X ∩ Y ∩ Z|
= 100 – |Wc ∪ Xc ∪ Yc ∪ Zc|
≥ 100 – |Wc| – |Xc| – |Yc| – |Zc|
= 100 – 35 – 20 – 20 – 10
= 15
So 15/100 = 15% is the minimal percentage, and we can reach this by making the complement sets disjoint (which is what we did in method 1).
Reference
100 Soldiers riddle – Mathematics Stack Exchange
https://math.stackexchange.com/questions/102598/100-soldiers-riddle
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