At least one working key on the ring

Question

A key ring has 8 keys, of which 2 open the front door and 6 do not. You pick 3 keys at random to try. What is the probability that at least one of your 3 keys opens the door?

Accepted Answer

Counting "at least one working key" directly would mean adding the cases of exactly 1 and exactly 2. It is easier to find the chance of "no working key" and subtract from 1. Chance of zero working keys means all 3 come from the 6 useless keys. With combinations, where \binom{n}{k} counts ways to choose k from n: P(	ext{none work})=\frac{\binom{6}{3}}{\binom{8}{3}}=\frac{20}{56}=\frac{5}{14}. Subtract from 1: 1-\frac{5}{14}=\frac{9}{14}\approx 0.643. Takeaway. The complement turns a messy "at least one" into one tidy count of "none," then a subtraction. This is the go-to move whenever "at lea…

At least one working key on the ring

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