Choosing numbers at random - expected value calculation From set $\{1,2,\ldots,49 \}$ we choose at random 6 numbers without replacing them. Let X denotes quantity of odd numbers chosen. Find $\mathbb{E}X$, how to find that? I have no idea whatsoever. 
EDIT:: still looking for the sufficient explanation. 

We consider the $7$ cases and determine how many odd balls are present in each case.
The $7$ cases are:


*

*$6$ even : $\dbinom{24}{6}$

*$5$ even, $1$ odd : $\dbinom{24}{5}\cdot 25$

*$4$ even, $2$ odd : $\dbinom{24}{4}\cdot \dbinom{25}{2}$

*$3$ even, $3$ odd : $\dbinom{24}{3}\cdot \dbinom{25}{3}$

*$2$ even, $4$ odd : $\dbinom{24}{2}\cdot \dbinom{25}{4}$

*$1$ even, $5$ odd : $24\cdot \dbinom{25}{5}$

*$6$ odd : $\dbinom{25}{6}$


Calculate all of these, multiply each by the corresponding number of odd balls, for example, $\dbinom{24}{4}\cdot \dbinom{25}{2}\times 2$, add them all up and divide by the total number of cases, $\dbinom{49}{6}$
