# »  Solutions to puzzles in my VDARE.com monthly Diary

## November 2022

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In the Math Corner of my November 2022 Diary I invited readers to try a question from the gāokăo, China's equivalent of the SAT. I noted:

We hear that the gāokăo is ferociously difficult. Looking at some of the math questions just now, though, I thought I would have coped easily with them at age 18. When I said that to Mrs Derbyshire, however, she jeered at me as a math snob. Eh, maybe.

See what you think. Here, as a brainteaser, is one I took from an article about this year's gāokăo. (The article, I should warn, includes a worked solution. Don't cheat — it's illegal!)

Brainteaser:  Four students can choose to participate in a charity event either on Saturday or Sunday but not both. What's the probability that there is at least one student at both Saturday's and Sunday's events? Is it 1/8, 3/8, 5/8, or 7/8?

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•  Solution

The probability that there is at least one student at both Saturday's and Sunday's events is the inverse of, i.e. 1 minus, the probability that there is no student at either Saturday's event or Sunday's.

The probability that all four students stayed home on Saturday, assuming it's decided by a coin toss, is 1/16. Same for Sunday. The probability that all four students stayed home on either Saturday or Sunday is therefore 2/16, i.e. 1/8. The inverse is 7/8, so that's the solution.

If you want to get fancy you can expand (½ + ½) by the binomial theorem. That gives 1/16 + 1/4 + +3/8 + 1/4 + 1/16. Each term there corresponds to the Saturday-Sunday probability of 4-0, 3-1, 2-2, 1-3, 0-4 students present.

For the sake of the gāokăo's reputation, I hope at least some of it is harder than that.