Is my calculation correct about a probability problem? Suppose there are $300$ tickets in the pool, where $7$ of them belong to me. $20$ tickets are randomly taken out of the pool, and are declared as "winning tickets". What is the probability that exactly 4 of the winning tickets are mine?
When I tried to solve this I found
$$\frac{\binom{20}{4} \left(7 \times 6 \times 5 \times 4 \prod _{j=1}^{16} (-j+293+1)\right)}{\prod _{i=1}^{20} (-i+300+1)} \approx 0.000433665 $$
Is this the correct probability?
Thanks.

Your answer is correct but the way it is notated is not very elegant. 
I should choose to write it as:

$$\frac{\binom{20}{4}\binom{280}{3}}{\binom{300}{7}}$$

Choosing $7$ from $300$ gives $\binom{300}{7}$ possibilities. $4$ of them belonging to your $20$ gives $\binom{20}{4}$ possibilities and $3$ of them belonging to the $280$ that are not yours gives $\binom{280}{3}$ possibilities.
