Urn has 5 black and 5 white balls. X is RV that is number of balls you draw till first white. Calculate the PMF, E[X] and Var(X) An urn contains 5 black and 5 white balls. You draw without replacement till you draw your first white. X is a RV representing the total balls you have drawn including the first white.  
a) What is the PMF?
$5\choose1$/$10\choose 1$ for $1\le X<2$
$5\choose1$$5\choose1$/$10\choose2$ for $2 \le X <3$
$5\choose 2$$5\choose 1$/$10\choose3$ for $3 \le X < 4$
$5\choose 3$$5\choose 1$/$10\choose4$ for $4 \le X < 5$
$5\choose 4$$5\choose 1$/$10\choose5$ for $5 \le X < 6$
$5\choose 5$$5\choose 1$/$10\choose6$ for $6 \le X < 7$
b)What is E[X]?
$5\choose1$/$10\choose 1$+2[$5\choose1$$5\choose1$/$10\choose2$]+3[$5\choose 2$$5\choose 1$/$10\choose3$]+4[$5\choose 3$$5\choose 1$/$10\choose4$]+5[$5\choose 4$$5\choose 1$/$10\choose5$]+6[$5\choose 5$$5\choose 1$/$10\choose6$]
c)What is Var(X)?
$E[X]^2-E[X^2]$
$E[X]^2-$$5\choose1$/$10\choose 1$+2[$5\choose1$$5\choose1$/$10\choose2$]+3[$5\choose 2$$5\choose 1$/$10\choose3$]+4[$5\choose 3$$5\choose 1$/$10\choose4$]+5[$5\choose 4$$5\choose 1$/$10\choose5$]+6[$5\choose 5$$5\choose 1$/$10\choose6$]
Is this all correct? I feel unconfident about the PMF, but that's what everything else relies on.
 A: You have not accounted for the order of selection.  If the first white ball is drawn on the $k$th draw, then black balls must be drawn during the first $k - 1$ draws.  The probability that this occurs is found by multiplying the probability that black balls are drawn in each of the first $k - 1$ draws by the probability that the $k$th ball drawn is white given that the first $k - 1$ balls were black.  The probability that the first $k - 1$ balls drawn are black is 
$$\frac{\dbinom{5}{k - 1}}{\dbinom{10}{k - 1}}$$
since each of the first $k - 1$ balls must be drawn from among the five black balls when drawing $k - 1$ balls from the $5 + 5 = 10$ balls in the urn.  The probability that the $k$th ball drawn is white given that the first $k - 1$ balls were black is 
$$\frac{5}{10 - (k - 1)} = \frac{5}{11 - k}$$
since $5$ of the remaining $10 - (k - 1) = 11 - k$ balls are white.  Therefore,
the probability that the first white ball is drawn on the $k$th draw, where $1 \leq k \leq 6$, is
$$\Pr(X = k) = \frac{\dbinom{5}{k - 1}}{\dbinom{10}{k - 1}} \cdot \frac{5}{11 - k}$$
