First, we need the probability that among the first five balls, excactly three were black (and therefore two were white). This is a standard hypergeometric probability exercise, and the answer is
$$
\frac{\binom53\binom72}{\binom{12}5} = \frac{35}{132}
$$
Then, the probability that after this, the sixth ball is white. At this point there are seven balls left, and five of them are white. So the probability of drawing one of them as the seventh ball is
$$
\frac{5}{7}
$$
Now just multiply these two probabilities together, and you get your $\frac{25}{132}$.
There are $5$ black and $15$ white balls in a box. Balls of the same color are indistinguishable. We randomly remove balls from the box without replacement and arrange them in a row. Calculate the probability that all the black balls are adjacent.
Since balls of the same color are indistinguishable, two arrangements are distinguished by which five of the $20$ positions in the row are occupied by black balls. Hence, there are $\binom{20}{5}$ possible arrangements in the sample space, as Thomas Andrews stated in the comments.
If all five black balls are consecutive, the block of black balls must begin in one of the first $16$ positions.
Hence, the probability that all the black balls appear consecutively is
$$\frac{16}{\dbinom{20}{5}}$$
Calculate the probability that between every two black balls there will be at least two white balls.
The complementary event to the event that there are at least two white balls between each pair of successive black balls is that there is at most one white ball between each pair of successive black balls.
Place the five black balls in a row. This creates six spaces in which to place the white balls, four between successive black balls and two at the ends of the row.
$$\square b \square b \square b \square b \square b \square$$
Let $x_i$ be the number of white balls placed in the $i$th such space. Then
$$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 15 \tag{1}$$
is an equation in the nonnegative integers subject to the constraints that $x_2, x_3, x_4, x_5 \ge 2$ since there must be at least two white balls between any pair of successive black balls.
Let $x_2' = x_2 - 2$, $x_3' = x_3 - 2$, $x_4' = x_4' - 2$, and $x_5' = x_5 - 2$. Then $x_2', x_3', x_4', x_5'$ are nonnegative integers. Substituting $x_2' + 2$ for $x_2$, $x_3' + 2$ for $x_3$, $x_4' + 2$ for $x_4$, and $x_5' + 2$ for $x_5$ in equation $1$ and simplifying yields
$$x_1 + x_2' + x_3' + x_4' + x_5' + x_6 = 7 \tag{2}$$
Equation 2 is an equation in the nonnegative integers. A particular solution of equation $2$ corresponds to the placement of $6 - 1 = 5$ addition signs in a row of $7$ ones. For instance,
$$1 1 + + 1 + 1 1 1 + 1 +$$
corresponds to the solution $x_1 = 2, x_2 = 0, x_3 = 1, x_4 = 3, x_5 = 1, x_6 = 0$. The number of solutions of equation $2$ in the nonnegative integers is the number of ways five addition signs can be placed in a row of seven ones, which is
$$\binom{7 + 6 - 1}{6 - 1} = \binom{12}{5}$$
since we must choose which five of the $12$ positions required for seven ones and five addition signs will be filled with addition signs.
Hence, the probability that there are at least two white balls between each pair of black balls is
$$\frac{\dbinom{12}{5}}{\dbinom{20}{5}}$$
Calculate the probability that the third ball in the arrangement is a black ball.
In your attempt, you did not account for the fact that balls of the same color are indistinguishable.
Place a black ball in the third position. That leaves four black balls and $15$ white balls to be distributed to the remaining $19$ positions, which can be done in $\binom{19}{4}$ ways. Hence, the probability that there is a black ball in the third position is
$$\frac{\dbinom{19}{4}}{\dbinom{20}{5}} = \frac{\dfrac{19!}{4!15!}}{\dfrac{20!}{5!15!}} = \frac{19!}{4!15!} \cdot \frac{5!15!}{20!} = \frac{19!}{4!} \cdot \frac{5!}{20!} = \frac{19!}{4!} \cdot \frac{5 \cdot 4!}{20 \cdot 19!} = \frac{5}{20} = \frac{1}{4}$$
There is an easier way to see this. There are $20$ balls, of which $5$ are black and $15$ are white. If we draw a ball at random and place it in the third position, $5$ of the $20$ balls we could place there are black. Hence, the probability that a black ball is in the third position is
$$\frac{5}{20} = \frac{1}{4}$$
Best Answer
Here is a recursive method:
Let $P(n)$ denote the answer for a string of length $n$ . We remark that $$n≤3\implies P(n)=1 \quad \&\quad P(4)=1-\left(\frac 23\right)^4$$
For $n>4$ we remark that if a sequence is good then it must begin with exactly one of $B,WB,W^2B,W^3B$ and of course it must then be followed by a good sequence of shorter length. It follows that $$P(n)=\frac 13\times P(n-1)+\frac 23\times \frac 13\times P(n-2)+\left( \frac 23\right)^2\times \frac 13\times P(n-3)+\left( \frac 23\right)^3\times \frac 13\times P(n-4)$$
This is very easy to implement mechanically, maybe a bit slow to do with pencil and paper.
We get $$P(5)=0.736625514,\quad P(6)=0.670781893,\quad\boxed {P(7)=0.604938272}$$
Sanity Check: Let's get $P(5)$ by hand. The only "bad" sequences of length $5$ are $W^4B, BW^4, W^5$. Easy to compute the probabilities of each and we get $$P(5)=1-2\times \left(\frac 23\right)^4\times \frac 13-\left( \frac 23 \right)^5=0.736625514$$ which matches the result obtained by the recursion.