Urn 1 has 1 orange ball and 6 blue balls. Urn 2 has 2 orange balls and 5 blue balls. Suppose you draw 3 balls from one urn. To decide which urn to use you roll a fair 6-sided die. Draw from urn 1 if you roll an even number, urn 2 if you roll an odd number. What's the probability of drawing exactly one orange ball?
I understand that you have a $0.5$ chance of drawing from urn 1 and $0.5$ chance of drawing from urn 2. I initially drew out a tree diagram for this question which led me to the answer of $P(exactly\ 1\ orange) = 0.5(3/7 + 4/7)$. My issue is the other solution which involves combinations.
$$P(1\ orange | Urn_1) = \frac{6 \choose 2}{7 \choose 3} = 15 / 35 = \frac{3}{7}$$ and $$P(1\ orange | Urn_2) = 2 \left(\frac{5 \choose 2}{7 \choose 3} \right) = 20/35 = 4/7$$
My mind simply can't understand why the above works. I have a tree diagram in front of me where I manually calculate each of the options but I can't relate the two together.
I do know in the end you would just do $$P(1\ orange) = 0.5 \left(P(1\ orange | Urn_1) + P(1\ orange | Urn_2) \right) = 0.5$$