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.

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 $\frac{7}{16}$. 
