Distribution of amount of green balls drawn Urn of type A contains 4 green and 3 blue balls. Urn of type B contains 4 blue and 5 green balls.
There are three urns of type A and two urns of type B. We pick one urn at random (out of 5) and we draw 3
balls (with a single draw) out of the chosen urn. Let X denote the number of green balls drawn. Determine
the distribution of X.
I tried the following approach
$
P(X=N)=\frac{3}{5}\binom{3}{N}\frac{\binom{4}{N}\binom{3}{3-N}}{\binom{7}{3}}+\frac{2}{5}\binom{3}{N}\frac{\binom{5}{N}\binom{4}{3-N}}{\binom{7}{3}}
$
where $N \in \{ 0,1,2,3 \}$
However sum of probabilities is greater than 1:
$
\sum_{N=0}^{3} \frac{3}{5}\binom{3}{N}\frac{\binom{4}{N}\binom{3}{3-N}}{\binom{7}{3}}+\frac{2}{5}\binom{3}{N}\frac{\binom{5}{N}\binom{4}{3-N}}{\binom{7}{3}} = \frac{283}{105}
$
I have a problem with understanding how to calculate the probability of obtaining N green balls from a single draw.

I think you may have some unneeded terms.  You want $\frac35$ times some hypergeometric probability and $\frac25$ times some hypergeometric probability as you seem to have found.  But then you included $\binom{3}{N}$ which I do not understand.  You also had a $7$ instead of a $9$ in the bottom right denominator.
Perhaps $$\mathbb P(X=N)=\frac{3}{5}\frac{\binom{4}{N}\binom{3}{3-N}}{\binom{7}{3}}+\frac{2}{5}\frac{\binom{5}{N}\binom{4}{3-N}}{\binom{9}{3}}$$ might sum to $1$.
