# If 2 out of 6 Machines are Defective, and 3 were selected randomly , What is the Probability Distrib

If 2 out of 6 Machines are Defective, and 3 were selected randomly , What is the Probability Distribution of the number of Defectives?
Even more generally :
If n objects are divided to m alike and k alike , we select x amount randomly , then to find how probable is it that y objects from x are m , using some trial and error I managed to find this complicated formula:
${P}_{X}\left(y\right)=\frac{\left(\genfrac{}{}{0}{}{x}{y}\right){\left(}_{m}{P}_{y}\right){\left(}_{k}{P}_{x-y}\right)}{{}_{n}{P}_{x}}$
But then it turned out that there is an easier formula with Combinations which is actually exactly equal to the previous one
${P}_{X}\left(y\right)=\frac{\left(\genfrac{}{}{0}{}{m}{y}\right)\left(\genfrac{}{}{0}{}{k}{x-y}\right)}{\left(\genfrac{}{}{0}{}{n}{x}\right)}$
I can't understand how any of them can be shown mathematically or why do they have the same meaning.
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Explanation:
Suppose a bag contains 8 Red and 4 Blue balls. 3 balls are chosen at random. Find the probability that 2 of the 3 are Red.
1. Total number of different choices of 3 from 12 =$\left(\genfrac{}{}{0}{}{12}{3}\right)=220$
2. Number of ways of choosing 2 Reds from $8=\left(\begin{array}{c}8\\ 2\end{array}\right)$.
3. Number of ways of choosing 1 Blue from $4=\left(\begin{array}{c}4\\ 1\end{array}\right)$
4. Hence total number of ways of choosing 2 Reds and 1 Blue =$\left(\genfrac{}{}{0}{}{8}{2}\right)\left(\genfrac{}{}{0}{}{4}{1}\right)=112$
5. Hence probability of choosing 2 Reds and 1 Blue =$\frac{\left(\genfrac{}{}{0}{}{8}{2}\right)\left(\genfrac{}{}{0}{}{4}{1}\right)}{\left(\genfrac{}{}{0}{}{12}{3}\right)}=\frac{112}{220}$.