I saw this problem a while ago and a found the same answer as the textbook, yet I wasn’t truly sure if that was correct. Here was the problem:
You have a wheel with 4 equal sections, red, yellow, green and blue. Another wheel has 3 equal sections: red, yellow, and green. Knowing that exactly 1 wheel landed green, what are the odds the second wheel was the one that did so.
I immediately counted: if the first when is green, there are 2 possibilities for the second wheel: red and yellow. If however the second wheel landed green, there are 3 possibilities for the first wheel. Red, yellow and blue. I hence concluded there are 3 out of 5 possibilities in which the second wheel was green, so the answer is 3/5. This however doesn’t seem correct as we didn’t account for the fact that the first wheel has 4 sections, and the second has 3. Am I correct, or is there something I didn’t account for? Thank you in advance!
The problem is small so we can brute force it by counting to have certainty in the calculation.
The possible outcomes are (using 1-4 for colors, wheel 1 then wheel 2):
11
12
13
21
22
23
31
32
33
41
42
43
Of these, there are 5 spins with exactly one green (#1), and of these 3 of them are in the second wheel, 2 in the first, so the odds are 3:2. The probability is $\frac{3}{5}=0.6$
But then brute force is usually not possible, and the more compact route to the solution would be:
Probability of green only in wheel 2: $\frac{3}{4}\frac{1}{3}=\frac{1}{4}$
Probability of green only in wheel 1: $\frac{1}{4}\frac{2}{3}=\frac{1}{6}$
Odds: $\frac{\frac{1}{4}}{\frac{1}{6}}=\frac{3}{2}$
Probability: $\frac{\frac{1}{4}}{\frac{1}{6}+\frac{1}{4}}=\frac{\frac{1}{4}}{\frac{2}{12}+\frac{3}{12}}=\frac{12}{20}=\frac{3}{5}$