There are $$\binom{100}{r}$$ sequences of length $100$ consisting of $r$ red balls and $100-r$ non-red balls.
The number of sequences in which the first ball is red is $\binom{99}{r-1}$. The number of sequences in which the 50-th ball is red is $\binom{99}{r-1}$. The number of sequences in which the 100-th ball is red is $\binom{99}{r-1}$.
In fact, for any $i \in \{1,2,\ldots,100\}$, the number of sequences in which the $i$-th ball is red is $\binom{99}{r-1}$.
Hence, for any $i \in \{1,2,\ldots,100\}$, the probability that the $i$-th ball is red is $$\frac{\binom{99}{r-1}}{\binom{100}{r}}=\frac{r}{100}.$$
Hint: We can model this by arranging the ten balls in a line, so that the first four are those moved into the second urn then removed in that order. Given that the first ball is white, what is the probability that the second ball is too?
That is the short method.
The long method is:
Let $E_x$ be the event of $x$ white balls among the four balls drawn from five white and five black, for $x\in\{0,1,2,3,4\}$. This corresponds to a hypergeometric distribution.
$$\mathsf P(E_x) ~=~ \dfrac{\dbinom 5 x\dbinom 5 {4-x}}{\dbinom {10}4}\cdot\mathbf 1_{x\in\{0,1,2,3,4\}}$$
The events $E_0,E_1,E_2,E_3,E_4$ are thus not equally probable.
Let $A, B$ be the event of drawing white balls in two consecutive draws from among those four.
$$\mathsf P(B\mid A)=\dfrac{\sum_{x=2}^4 \mathsf P(E_x)~\mathsf P(A\cap B\mid E_x)}{\sum_{x=1}^4 \mathsf P(E_x)~\mathsf P(A\mid E_x)}=\dfrac{\sum_{x=2}^4 \binom 5x\binom 5{4-x}\binom x 2/\binom 4 2}{\sum_{x=1}^4 \binom 5x\binom 5{4-x}\binom x 1/\binom 4 1}$$
Best Answer
If you want to complicate matters, you could compute
$\text{(ways of placing 59 black balls in 99)/(ways of placing 100 black balls in 60)}= \dfrac{\binom{99}{59}}{\binom{100}{60}}$
but the book's suggestion is good. Imagine all the balls to be randomly placed in a line, and ask
what is the probability that the first ball, say, is black ? $\frac35$
what is the probability that the ninth ball, say, is black ? $\frac35\;\;$
what is the probability that the $i^{th}$ ball is black ? $\frac35\;\;$ isn't it ?
so what is the probability that the last ball is black ?