If we extract $3$ marbles randomly from an urn that contains... If we extract $3$ marbles randomly from an urn that contains $6$ white marbles and $5$ black marbles, what is the probability that one is white and the other two black?

Solution. If we take into account the order of extraction and suppose that every possible outcome of the sample space is equiprobable, the probability required is

$\frac{(6 · 5 · 4) + (5 · 6 · 4) + (5 · 4 · 6)}{11 · 10 · 9}=\frac{3 · 120}{990}=4/
11= 0.3636$

I don't understand the numerator $(6 · 5 · 4) + (5 · 6 · 4) + (5 · 4 · 6)$. The denominator $11 · 10 · 9$ are the possible cases, you have $11$ possibilities for the first marbles, $10$ for the second, $9$ for the last one.
There are $3$ mutual exclusive possibilities: WBB, BWB, BBW.

This with: 


$P(WBB)=\frac6{11}\frac5{10}\frac4{9}$
$P(BWB)=\frac5{11}\frac6{10}\frac4{9}$
$P(BBW)=\frac5{11}\frac4{10}\frac6{9}$


More precise we can write: $$P(BWB)=P(B_1\cap W_2\cap B_3)=P(B_1)P(W_2\mid B_1)P(B_3\mid B_1\cap W_2)=\frac5{11}\frac6{10}\frac4{9}$$where e.g. $W_2$ stands for the event that the second chosen marble is white.

Addition of these probabilities gives you the probability of the mentioned event.
They are considering the three possible orders of selecting one white and two blacks: $WBB, BWB$ and $BBW$. E.g. for $WBB$ (first marble you pick is white, then next two are black), the number of ways to do this is $6\times 5\times 4$, since you can pick the white marble in $6$ ways, then pick a black marble in $5$ ways, then pick the second black marble in $4$ ways.