It's kind of hard to calculate this directly but of course there is a trick! We're going to calculate the chance that neither cop is a domestic abuser instead. Then all other sets of two cops will contain at least one

There's a 60% chance that cop one isn't and a 60% chance that cop two isn't. This makes the chance that both aren't 0.6 x 0.6 = 0.36. so 36% of pairs of cops fail to contain a domestic abuser, which means 64% contain at least one. If you see two cops and say that there's a 64% chance that one is an abuser, well, that is correct!

This keeps working! The chance that none in a set of three is an abuser is 0.6 x 0.6 x 0.6 = (0.6)^3 = 0.216, so there's a 78.4% = 100-21.6 chance that a set of three cops contains at least one abuser.

If there are n cops the formula is:

(1-(0.6)^n)*100

n. % chance of at least one
1. 40
2. 64
3. 78.4
4. 87
5. 92.2
6. 95.3

Word problems don't have to be counterrevolutionary...

#ACAB #PoliceAbolition #Abolition #WordProblems #DiscreteProbability #Mathstodon