Find the probabilities that the fourth White ball is the fourth, fifth, sixth or seventh ball drawn if the sampling is done without replacement An urn contains 10 Red and 10 White balls. The balls are drawn from the urn at random, one at a time. Find the probabilities that the fourth White ball is the fourth, fifth, sixth or seventh ball drawn if the sampling is done without replacement.
Here's what I've done so far:
$$\text{Let } B_k = \text{ Probability that the fourth White ball is the } k^{th} \text{ ball drawn}$$
$$P(B_4) = \frac{10 \choose 4}{20 \choose 4} = \frac{14}{323}$$
since clearly all 4 balls need to be white so are chosen from the 10 white options.
Now here is where I start having a problem:
For $P(B_5)$, I was thinking that obviously the fifth ball needs to be white so of the first 4 balls one can be red and the other 3 white.
$$P(B_5) = \frac{{10\choose 4} {10\choose 1} {4 \choose 1}}{20 \choose 5} = \frac{175}{323}$$
but I don't think this is right. The solution I've been given is $\frac{35}{323}$
What I thought was choose 4 White and 1 Red; then of the first 4 places choose 1 to be Red (which would be the same as choosing 3 White)
If I divide my answer by 5 I get the solution. I don't know why this is. I am also not very confident in my solution because I feel like there is something wrong with the ordering.
 A: This is the same as the probability that in the first $7$ draws there are $4$, $5$, $6$, or $7$ whites.
That probability is 
$$\frac{\binom{10}{4}\binom{10}{3}+\binom{10}{5}\binom{10}{2}+\binom{10}{6}\binom{10}{1}+\binom{10}{7}\binom{10}{0}}{\binom{20}{7}}.$$
Compute. There will be an awful lot of cancellation.
Remark: In your notation, $\Pr(B_5)$ is indeed $35/323$. For the fifth ball is the fourth white red if there are $3$ white in the first $4$ and then a white. The probability of $3$ white in the first $4$ is $\frac{\binom{10}{3}\binom{10}{1}}{\binom{20}{4}}$. Now multiply by $\frac{7}{16}$. 
A: The tricky part here is that for the first four balls, it doesn't matter the ordering, as long as you get 3 white and 1 red.  But the fifth has to be white.  Your computation mixes a bit of ordering with a bit of non-ordering.  I would probably do these separately: first, the probability of choosing three white and one red out of the first four is:
$$\frac{\left( \begin{array}{c} 10 \\ 3 \end{array} \right) \left( \begin{array}{c} 10 \\ 1 \end{array} \right)}{\left( \begin{array}{c} 20 \\ 4 \end{array} \right)}.$$
Now there are 16 balls left, and 7 are white, so multiplying the above expression by $7/16$ gives your answer.
By the way, here is the reason why you need to divide yours by $5$.  In your denominator is the total number of unordered ways to choose 5 balls from 20.  But your numerator takes into account some ordering (you are trying to put the red one as one of the first four).  If you include this ordering in the numerator, you must also account for it in the denominator.  You are allowing 4 spots for the red one out of five possible spots.  So you should not be multiplying by 4, but rather by $4/5$.
