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Let the hats of monkeys A, B and C be a, b and c respectively.
This table shows the ways in which the monkeys can select hats.
In only two of the six ways do none of the monkeys have the same hat as when they arrived, hence the required probability is
$$\frac{2}{6}=\frac{1}{3}$$
Monkey Puzzle
Age 11 to 14
ShortChallenge Level 
Let the hats of monkeys A, B and C be a, b and c respectively.
This table shows the ways in which the monkeys can select hats.
| A | B | C |
|---|---|---|
| a | b | c |
| a | c | b |
| b | a | c |
| b | c | a |
| c | a | b |
| c | b | a |
In only two of the six ways do none of the monkeys have the same hat as when they arrived, hence the required probability is
$$\frac{2}{6}=\frac{1}{3}$$
This problem is taken from the UKMT Mathematical Challenges.
You can find more short problems, arranged by curriculum topic, in our short problems collection.
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