Probability of presence of not defective products in a pack. In a pack of 50 products there mustn't be more than 2 defective products.
The probability of presence of 0, 1 or 2 defective products in the pack is the same.
Five randomly chosen products are not defective.
What's the probability of the rest of the products in the pack being not defective too?

My attempt to solution:
$$
   P_0 = P_1 = P_2 = \frac{1}{3} \\
   P_0 + P_1 + P_2 = 1
$$
Where $ P_i $ is the probability of presence of i defective products in the pack with 50 products.
I assumed that there could be present conservation law (hence the second equation).

The 5 selected products that are taken out of the pack form $ \frac{5}{50} = \frac{1}{10}\text{th} $ of the initial number of products in the pack.

If the 5 selected products are not defective, the $ P_{0} $ must have decreased in favour of $ P_1 \text{ and } P_2$ (the conservation law).

The question is: what's the new probability $ P'_{0} $ of no defective products being present in the pack?

Based on previous assumptions, I thought that $ P'_{0} = \frac{1}{3} - (\frac{1}{10}\cdot\frac{1}{3} ) = \frac{3}{10} $.
Let's denote the events of $0$, $1$, or $2$ defectives by $D_0$, $D_1$, $D_2$; and the event of getting $5$ goods by $G$.

We are interested in the conditional probability $P(D_0 \mid G)$.

We have
$$P(D_0 \mid G)=\frac{P(D_0\cap G)}{P(G)}$$
Noting that $D_0\subseteq G$, we obtain
$$P(D_0 \mid G)=\frac{P(D_0)}{P(G)}$$
We know $P(D_0)=\frac13$.

We can also calculate 
$$ \begin{array}{lcl}P(G)&amp;=&amp;P(G\mid D_0)\cdot P(D_0)+P(G\mid D_1)\cdot   P(D_1)+P(G\mid D_2)\cdot P(D_2)\\
&amp;=&amp;1\cdot \frac13+\frac{9}{10}\cdot\frac13+\frac{48\choose 5}{50\choose 5}\cdot \frac13
\end{array}$$
You should be able to finish it from there.
Let's denote the events of $0$, $1$, or $2$ defectives by $D_0$, $D_1$, $D_2$; and the event of getting $5$ goods by $G$.

We are interested in the conditional probability $P(D_0 \mid G)$.

We have
$$P(D_0 \mid G)=\frac{P(D_0\cap G)}{P(G)}$$
Noting that $D_0\subseteq G$, we obtain
$$P(D_0 \mid G)=\frac{P(D_0)}{P(G)}$$
We know $P(D_0)=\frac13$.

We can also calculate 
$$ \begin{array}{lcl}P(G)&amp;=&amp;P(G\mid D_0)\cdot P(D_0)+P(G\mid D_1)\cdot   P(D_1)+P(G\mid D_2)\cdot P(D_2)\\
&amp;=&amp;1\cdot \frac13+\frac{9}{10}\cdot\frac13+\frac{48\choose 5}{50\choose 5}\cdot \frac13
\end{array}$$
You should be able to finish it from there.