Three boxes of coins
Three boxes each hold two coins: one box has two gold coins, one has two silver coins, and one has one gold and one silver coin. You pick a box at random and take out one of its two coins at random. It is gold.
What is the probability that the other coin in the same box is also gold?
Fractions, decimals with at least 3 significant digits, and expressions such as sqrt(2)/2, e^-1 or pi/4.
Count coins, not boxes. Which gold coin could you have drawn?
Use Bayes' rule: how likely is a gold first coin from each box?
h
s Shown after a correct answer. Revealing it first marks the problem as seen, not solved.
Idea. The gold coin you drew is equally likely to be any of the three gold coins. Two of those three sit in the box with two gold coins.
By counting. Every coin is equally likely to be the one drawn: each box has probability and each coin in it . Given that the coin is gold, it is one of the three gold coins, each equally likely. Two of them are in the gold-gold box, where the other coin is gold; one is in the mixed box, where the other coin is silver. So the probability is .
By Bayes' rule. A gold coin comes out of the gold-gold box with probability 1, of the mixed box with probability , of the silver box with probability 0. So
The trap. “The box is either the gold-gold one or the mixed one, so it is .” The two boxes are not equally likely once you know you drew gold: the gold-gold box produces gold twice as often.