Consider two urns with the following composition: Urn $1$ contains $7$ red
and $3$ black balls, Urn $2$ contains $4$ red and $6$ black balls. Our random
experiment consists of choosing one of the two urns with equal
probability and then drawing a sample (without replacement) of $3$
balls. If $3$ red balls are drawn, what is the probability they came
from Urn $1$?
Let $R$ be the event that three red balls are drawn without replacement. Let $U1$ be the event that Urn $1$ is chosen. Then by Bayes theorem, $$P(U1|R) = \frac{P(R|U1)P(U1)}{P(R)}$$ where
$$P(R|U1) = (7/10)(6/9)(5/8)$$ and
$$P(U1) = 0.5$$
Edit: I badly misread the question, the sample is three at a time.
Best Answer
Let $~\displaystyle p_1 ~\text{denote} ~\frac{\binom{7}{3}}{\binom{10}{3}}.$
Let $~\displaystyle p_2 ~\text{denote} ~ \frac{\binom{4}{3}}{\binom{10}{3}}.$
Then, in accordance with Bayes Theorem, the probability that the balls came from Urn 1 is
$$\frac{p_1}{p_1 + p_2} = \frac{\binom{7}{3}}{\binom{7}{3} + \binom{4}{3}}.$$