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I'm trying to understand how to calculate probability that one of drawn elements is of certain type, for example:

$|\Omega| = 40$ where $24$ elements are of type $X$ and $16$ are of type $Y$, so $$\Omega = \{X_{1},\dots,X_{i},Y_{1},\dots, Y_{j};i=24;j=16\}$$

You draw $7$ elements without replacement. What is the probability that at least $1$ of drawn elements is of type $X$?

I know answer for drawing a single element, then it is just $\frac{24}{40}$, but for multiple elements I have completely no clue.

Also let me know if my question is hard to understand, English is not my first language.

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    Are you drawing with or without replacement? – 2017-01-06
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    I updated my question – 2017-01-06
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    In either case, it is easier to compute the probability that your draw only contains type $Y$ (then subtract from $1$). With replacement that's $\left(\frac {16}{40}\right)^7$. Without replacement it's $\frac {16}{40}\times \frac {15}{39}\times\cdots \times \frac {10}{34}$. – 2017-01-06
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    As an alternate way to think about it, there are $\binom {40}{7}$ ways to draw $7$ elements. There are $\binom {16}{7}$ ways to draw $7$ elements of type $Y$, so just compute the ratio (and subtract from $1$). – 2017-01-06
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    @lulu What would happen if I wanted to calculate probability that my draw contains exactly 3 elements of type X? – 2017-01-06
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    In that case you need to choose your three $X$ elements, $\binom {24}{3}$, then choose the four $Y$ elements, $\binom {16}{3}$. Multiply and divide by the total $\binom {40}7$. – 2017-01-06
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    @lulu Thank you very much. I understand it now! You can write it as an answer, so I can accept it. – 2017-01-06

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A good general way to approach these problems is to count the number of combinations which either do or do not have the desired property.

For the posted question, it's easiest to count the number of combinations that do not contain any $X$ elements. That's $\binom {16}7$. As there are $\binom {40}7$ unrestricted combinations the answer you want is the ratio $$\frac {\binom {40}7\binom {16}7}{\binom {40}7}=1- \frac {\binom {16}7}{\binom {40}7}=\frac {35831}{35853}\sim 0.99938638$$

In the comments you asked about the probability that your draw had exactly three $X$ elements. My answer there had a typo, the correct answer should be $$\frac {\binom {24} 3 \times \binom {16}4}{\binom {40}7}=\frac {7084}{35853}\sim 0.19758458$$