2 urns, blue and orange balls 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
$$P(1\ orange) = 0.5 \left(P(1\ orange | Urn_1) + P(1\ orange | Urn_2) \right) = 0.5$$

I think the way you wrote the two cases makes it harder to see the rule.  If you are drawing three balls, and want exactly $1$ of them to be orange, then the other two must be blue.  When drawing from an urn with $N_o$ orange balls and $N_b$ blue balls, then, you can choose the single orange ball in $N_o$ ways and the blue balls in ${N_b}\choose{2}$ ways; to obtain the probability, the product of these should be divided by the total number of ways to choose $3$ balls, which is ${N_o+N_b}\choose{3}$:
$$
P=\frac{N_o\cdot{{N_b}\choose{2}}}{{N_o+N_b}\choose{3}}.
$$
More generally, if you wanted to find the probability of drawing exactly $n_o$ orange balls and $n_b$ blue balls when drawing $n_o+n_b$ without replacement, it would be
$$
P=\frac{{{N_o}\choose{n_o}}{{N_b}\choose{n_b}}}{{N_o+N_b}\choose{n_o+n_b}}.
$$
