Try it on paper first. A hint, the key idea and the worked solution are here when you need them.
A made-up college has two ways of teaching a course, A and B, each with students. Under method A, of weak starters and of strong starters pass. Method B has an overall pass rate of . Let be the number of weak starters among A's students. Write A's overall number of passes in terms of . Find the smallest whole number for which A's overall pass rate is below B's.
Simpson's paradox: a comparison holds inside every group but reverses when the groups are combined. An overall rate is a weighted average of the group rates: add the passes of all the groups, then divide by the total number in all the groups.
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Step 1. A's passes. .
Step 2. The condition. A's rate is below B's when , so and .
Step 3. Check. At , A has passes, exactly equal to B. At , A has , just below.
Answer: A's passes are ; the smallest is .
What it means. If more than a third of A's students are weak starters, A can lose overall while it may still win inside every group.
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