Ask for help with a probability problem. The answer in the exercise book is different from my idea.

The answer is correct. Lz: You must think so. For example, Party A, Party B and Party C take the hat in turn, and Party A takes the hat first. The probability that he gets the hat with the correct color is 1/3, so there are two hats left, and the probability that Party B gets the hat with the correct color is 1/2 ... The last one left is Party C, so it is 1/3× 1/.

This algorithm is wrong. The premise for B to get the correct color (blue) is that A doesn't get a blue hat, and the probability of A not getting blue is 2/3, so the probability of B getting blue is 2/3× 1/2= 1/3. ...

In fact, imagine this problem as the calculation process of a, b and c taking hats in order should be

(1/3) × (2/3×1/2 )× (2/3×1/2×1) =1/27 ... That is to say, the probability of each person getting the correct color hat.

I don't know if you understand, but you can leave me a message if you have any questions.