combinatorics: probability of picking elements I have $6$ elements $$\{a, b, c, d, e, f \}$$
I close my eyes and randomly pick $5$ elements. 
What is the chance of getting $a$ and $b$ in those $5$ elements?
 A: Firstly, what are the subsets which contain $a$ and $b$? These are the subsets where we take $a$ and $b$ and an arbitrary set of 3 elements from the remaining ones. How many of these are there?
Now, what if the total number of subsets?
Assuming each subset is equally likely, what is the probability that you have drawn one of the subsets containing $a$ and $b$?
A: Suppose we will be unhappy if one of $a$ or $b$ is missing from our collection of $5$. With probability $\dfrac{1}{6}$, $a$ will be missing. The same is true of $b$. So the probability we will be unhappy is $\dfrac{1}{3}$.
It follows that with probability $\dfrac{2}{3}$, both $a$ and $b$ are in our chosen set. 
A: I would think of it slightly differently. Instead of picking $5$ elements to keep, we pick $1$ element to discard. There are then obviously $6$ different ways we can get rid of $1$ object from the set of $6$, and $4$ of the ways leave sets containing both $a$ and $b$, so the probability that we have is both $a$ and $b$ is $\frac{4}{6} = \frac{2}{3}$
A: from $6$ possible choices four of of them contain $a$ and $b$ together
$$abcde,abcdf,abcef,abdef$$
one does not contain only $b$
$$acdef$$
other does not contain only $a$
$$bcdef$$ 
 that mean probability is $$\frac{4}{6}=\frac{2}{3}$$ 
A: You have a set of 6 elements: $\mathbf{U}=\{a,b,c,d,e,f\}$. From this you select a set of 5 elements: $\mathbf{U}'=\{u_1. u_2, u_3, u_4, u_5, u_6 \}$. Denote event $E_1 =  \textit{' select both a and b into'  } \mathbf{U}'$. Clearly $P(E_1)=1-P(E_1^{c})=1-P(\textit{select either just a, or just b, or neither into } \mathbf{U}'$. 
Since $|\mathbf{U}'|=5$, it is impossible to select neither $a$ nor $b$. The other two subsets are $\mathbf{U}''=\mathbf{U}  /\{a\}$ and $\mathbf{U}'''=\mathbf{U}  /\{b\}$ with $|\mathbf{U}''|=5=|\mathbf{U}'|=|\mathbf{U}'''|$. Since sampling is without replacement and both outcomes are equally likely, the calculation is the following:
$$
P(E_1^{c})=\frac{5}{6} \cdot \frac{4}{5} \cdots \frac{1}{2} + \frac{5}{6} \cdot \frac{4}{5} \cdots \frac{1}{2} = 2 \cdot \frac{1}{6}=\frac{1}{3}
$$ 
Obviously what you need is 
$$
P(E_1)=1-\frac{1}{3}=\frac{2}{3}
$$
