How could outcomes be equally likely? Hi I've just started learning probability but I have some questions.
When sloving a problem like "If 3 balls are “randomly drawn” from a bowl containing 6 white and 5 black balls, what is the probability that one of the balls is white and the other two black?" , they assume outcomes in the sample space are equally likely. However when considering picking 2 balls from 2 white and 1000000 black balls, how could (w,w) (w,b) be (b,b) (unordered) considered equally likely? 
In my opinion (b,b) are much more likely than (w,w) because there are way much black balls than white balls. 
Of course their numbers are considered when solving the probability, I still wonder   is it valid that (w,w) (w,b) be (b,b) be assumed equally likely.
Sorry for my bad english because i'm not a native speaker. Please excuse me. Thank you.
-edited
In a textbook, it says that "When the experiment consists of a random selection of k items from a set of n items, we have the flexibility of either letting the outcome of the experiment be the ordered selection of the k items or letting it be the unordered set of items selected. In the former case we would assume that each new selection is equally likely to be any of the so far unselected items of the set, and in the latter case we would assume that all (n k) possible subsets of k items are equally likely to be the set selected." I'm curious about emphasied part.
 A: You are correct in saying that the probability of pulling (b,b) is more likely. This is because there are indeed more ways to choose two black balls than two balls in which one is white. 
When they say "all" outcomes are equally likely, they are referring to each individual outcome in which two balls are withdrawn regardless of their color. This may become more clear if we consider a simple example...
For example, if we have $2$ green balls in a bag and $4$ red balls in a bag, we can imagine our bag as a set $\{g_1,g_2,r_1,r_2,r_3,r_4\}$. If we were to choose ONE ball randomly from the bag, this would mean we choose them in such a way that every ball has an equal probability of being chosen, despite its color. In order words, we are as likely to choose $g_1$ as we are to choose $g_2$ as we are to choose $r_1$ as we are to choose... i.e. 
$$P(g_1)=P(g_2)=P(r_1)=P(r_2)=P(r_3)=P(r_4)$$
However, if you wanted the probability of a green ball being chosen, now I must count the number outcomes that satisfy this criteria and divide it by the total number of outcomes. In effect,
$$P(g_1 \vee g_2)=\frac{2}{6}=\frac{1}{3}$$
Suddenly probability of choosing green is not the same as choosing red. But that is because initially, when we said each outcome was equally likely, we were referring to each individual outcome, which was a ball, regardless of its color. 
A: In this context the equiprobable outcomes must be looked at as 3-sets of (distinguishable) balls. 
Think of the balls as numbered: $1,2,3,4,5,6$ are white and $7,8,9,10,11$ are black. 
Then an outcome by drawing $3$ balls without replacement is e.g. $\{5,2,9\}$.
There are $\binom{6+5}3=165$ of such outcomes and each of them has probability $\frac1{165}$.
We can define $W$ as the number of white balls that is drawn and $B$ as the number of black balls that is drawn, and $W,B$ must then be looked at as random variables on the outcome space.
If  $\omega=\{5,2,9\}$ then $W(\omega)=2$ and $B(\omega)=1$.

Also it is possible to go for $3$-tuples here. 
Then there will be $11\times10\times9=990$ equiprobable outcomes.
