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A box contains 4 black marbles, 7 red marbles and 8 green marbles. If Leon draws out 3 marbles 1 at a time without replacement, what are the odds that he will draw out 3 black marbles?

The sample space =$19$.

Since they are not being replaced then the probabilities are depended since the sample space gets smaller.

Would it be the odds of $$\frac{P(\text{black marble})}{P(\text{not black})} \rightarrow\frac{P(\frac{4}{19})}{P(\frac{15}{19})}?$$

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    You don't seem to be considering the "$3$" anywhere. Makes a big difference! For example, if you replace $3$ with $5$ the answer is clearly $0$. – 2017-02-09
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    Hint: I always believe probability is easier to have intuition for than odds (maybe just habit). To get from one to the other, note that $Odds(A)=\frac {P(A)}{1-P(A)}$. for an event $A$. – 2017-02-09
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    So I take the odds of not getting 3 black marbles right? – 2017-02-09
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    The answer is the odds of that, yes. Personally, I find it easier to compute the probability of getting $3$ black marbles. The probability that the first one is black is $\frac 4{19}$. What is the probability that the second is also black? – 2017-02-09
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    $\frac{3}{18}$ since we are not replacing – 2017-02-09
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    No...since there are only $18$ left to choose from. – 2017-02-09
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    I think you might want to try some more basic exercises. Are you working from a text? – 2017-02-09
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    Sorry I meant the $\frac{3}{18}$, when you take the third black marble you get $\frac{2}{17}$. My bad* – 2017-02-09
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    Yes, that's what the posted solution tells you. But getting other people to do your homework for you isn't a very good way to learn the subject. – 2017-02-09

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No, the "sample space" is NOT the number 19. It is the set of all possible combinations of 3 marbles from this set of marbles, not the number of marbles.

In any case, there are, initially, a total of 19 marbles, 4 of which are black. The probability the first marble drawn is black is $\frac{4}{19}$. If that happens then there are 18 marbles left, 3 of which are black. The probability the second marble drawn is black is $\frac{3}{18}=\frac{1}{6}$. Then there are 17 marbles left, 2 of which are black. The probability the last marble drawn is black is $\frac{2}{17}$. The probability of drawing 3 black marbles is $\frac{4}{19}*\frac{1}{6}*\frac{2}{17}=\frac{4}{969}$.

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    Okay, I get it know, I was just reducing the denominator by one but forget to get rid of one marble in the numerator as well. Thank you – 2017-02-09