3 of 10 products are defective and 5 are selected at random. Probability that sample includes all defective products 
$3$ of $10$ products are defective. If $5$ products are selected at random, what is the probability that all sample contains all three defective products

I have already calculated $P(X=3)= \dfrac1{12}$ and $P(X\geq1)=\dfrac1{12}$, where $X=$"Product is defective". After that, I have formed expression of conditional probability
$$P(P=3｜P\geq1)= \dfrac{P(P(X=3)\,\mathrm{and}\,P(X\geq1))}{P(X>=1)}$$
This is where I am stuck. The solution is $$\dfrac{P(X=3)}{1-P(X=0)}$$ but I have no idea where that $P(X\geq1)$ disappears from numerator $P(P(X=3) and P(X\geq1))$.
Thanks for your help!
 A: The underlying probability distribution for your question is called the hypergeometric distribution.
Each product can be classified into exactly one of two groups:  the defective group, and the non-defective group.  Now suppose there are $D$ defective items, and $G$ non-defective items in your population, for a total of $D+G$ total items.
Next, let's think about how you can choose products from this population of items without replacement.  Say you choose $d$ defective items and $g$ non-defective items, where $0 \le d \le D$ and $0 \le g \le G$.  Then the total number of items you chose is $d+g$.  How many ways can you make such a choice?  Well, the number of ways to make your choice among just the defective items only is simply $$\binom{D}{d} = \frac{D!}{d!(D-d)!}.$$  And similarly, among just the non-defective items, there are $$\binom{G}{g}$$ such ways.  But since the number of ways to choose items from one group does not depend on the number of ways to choose items from the other group, the total number of ways to choose items from both groups is the product:  $$\binom{D}{d} \binom{G}{g}.$$  This number represents the ways to choose exactly $d$ defective items and $g$ non-defective items from the population, and thus counts the number of desired outcomes.
But how many ways are there to choose $d+g$ items from $D+G$ products in general, without regard to classification?  In other words, how many ways could we have chosen the same total of items including cases where we didn't get exactly $d$ defective and $g$ non-defective items?  This is just $$\binom{D+G}{d+g}.$$  So the probability that we get exactly $d$ defective and $g$ non-defective items is $$\frac{\binom{D}{d}\binom{G}{g}}{\binom{D+G}{d+g}}.$$
In your case, you have $D = 3$ defective products in your population, and $D+G = 10$ total products, so $G = 7$ non-defective products.  You are choosing $d+g = 5$ total items from this population.  But you want to determine the probability of getting $d = 3$ defective items (all of them) and therefore $g = 5-3 = 2$ non-defective items.  The resulting probability is $$\frac{\binom{3}{3}\binom{7}{2}}{\binom{10}{5}}.$$
A: First, make sure you define all terms carefully! For example, I suppose that by $X$ you mean the number of defective products in your sample, rather than the event "product is effective" .. which isn't even clear as an event: which product is effective?
So, if $X=3$ is the event of having all $3$ defective products in your sample of $5$, and you already calculated that to be $\frac{1}{12}$, then congratulations, you are done! But somehow I doubt that's it.  Indeed, you also say that $P(X \ge 1)=\frac{1}{12}$ ... which makes little sense, since clearly the probability of getting at least one defective product in your sample should be greater than the probability of getting all $3$ defective samples, since you could easily get $1$ or $2$ defective samples. So, something is wrong here .. or you mean something different by your $X$, and by $P(X=3)$ and $P(X \ge 1)$ yet!
Now, what does make sense is that $P(X \ge 1) = 1 - P(X=0)$, for the probability of getting at least one defective product in your sample is $1$ minus the probability of getting no defective product in your sample at all.
Anyway, to calculate the probability of getting $3$ defective products in your sample, simply see how many samples of $5$ have all defective products out of all possible samples of $5$.
Well, there are $10 \choose 5$ possible samples of $5$ out of $10$.
Further, how many samples of $5$ contain all $3$ defective products?  Such samples need $2$ more products besides the $3$ defective ones, so you need to pick $2$ out of the $7$ remaining ones. So, there are $7 \choose 2$ samples that include all $3$ defective products.
Hence, the probability that you have all $3$ defective products in a sample of $5$ is:
$$\frac{7 \choose 2}{10 \choose 5}$$
