Choosing numbers at random - expected value calculation

Question

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.

Answer 1

Let $X_1, X_2, X_3, X_4, X_5, X_6$ be $0$ or $1$ according to whether the $i$th choice is even or odd. So $X=X_1+\cdots+X_6$. You can now use linearity of expectation.

Answer 2

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, $\dbinom{49}{6}$