The answers are (a) $40/100$; (b) $40/100$; (c) $40/100$.
(a) Since balls tend to roll around, let us imagine instead that we have $100$ cards, with the numbers $1$ to $100$ written on them. The cards with numbers $1$ to $40$ are red, and the rest are blue.
The cards are shuffled thoroughly, and we deal the top card.
There are $100$ cards, and each of them is equally likely to be the top card. Since $40$ of the cards are red, it follows that the probability of "success" (first card drawn is red) is $40/100$.
(b) If we think about (b) the right way, it should be equally clear that the probability is $40/100$. The probability that any particular card is the fifteenth one drawn is the same as the probability that it is the first one drawn: all permutations of the $100$ cards are equally likely. It follows that the probability that the fifteenth card drawn is red is $40/100$.
(c) First, fifteenth, eighty-eighth, last, it is all the same, since all permutations of the cards are equally likely.
Another way: We look again at (a), using a more complicated sample space.
Again we use the card analogy. There are $\binom{100}{40}$ ways to decide on the locations of the $40$ red cards (but not which red cards occupy these locations). All these ways are equally likely.
In how many ways can we choose these $40$ locations so that Position $1$ is one of the chosen locations? We only need to choose $39$ locations, and this can be done in $\binom{99}{39}$ ways.
So the desired probability is
$$\frac{\binom{99}{39}}{\binom{100}{40}}.$$
Compute. The top is $\frac{99!}{39!60!}$ and the bottom is $\frac{100!}{40!60!}$. Divide. There is a lot of cancellation. We quickly get $40/100$.
Now count the number of ways to find locations for the reds that include Position $15$. An identical argument shows that the number of such choices is $\binom{99}{39}$, so again we get probability $40/100$.
Another other way: There are other reasonable choices of sample space. The largest natural one is the set of $100!$ permutations of our $100$ cards. Now we count the permutations that have a red in a certain position, say Position $15$.
The red in Position $15$ can be chosen in $40$ ways. For each of these ways, the remaining $99$ cards can be arranged in $99!$ ways, for a total of $(40)(99!)$. Thus the probability that we have a red in Position $15$ is
$$\frac{(40)(99!)}{100!},$$
which quickly simplifies to $40/100$. The same argument works for any other specific position.
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 $$\frac{\binom{99}{r-1}}{\binom{100}{r}}=\frac{r}{100}.$$
Best Answer
Denote $N=100$ (number of balls), $K=r$ (number of red balls);
$P_{N,K}(n,k) = \Pr$ ($k$ balls are red, when $n$ draws are made), where $0\le k \le n$.
Looking at Hypergeometric distribution, one can see that
$$ P_{N,K}(n,k) = \dfrac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}}. $$
Then the probability that the $(n+1)$th ball (where $n=1,2,...,N-1$) drawn will be red, is: $$ P_{n+1} = \sum\limits_{k=0}^{n} P_{N,K}(n,k)\cdot \dfrac{K-k}{N-n} $$ $$ = \sum\limits_{k=0}^{n} \dfrac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}}\cdot \dfrac{K-k}{N-n} $$ $$ = \sum\limits_{k=0}^{n} \frac{ \frac{K\cdot(K-1)!}{k!(K-1-k)!} \binom{N-K}{n-k}}{ \frac{N\cdot(N-1)!}{n!(N-1-n)!}} $$ $$ =\dfrac{K}{N} \sum\limits_{k=0}^{n} \dfrac{ \binom{K-1}{k} \binom{N-K}{n-k}}{\binom{N-1}{n}} =\dfrac{K}{N} \sum\limits_{k=0}^{n} P_{N-1,K-1}(n,k)= \dfrac{K}{N} \cdot 1 = \dfrac{K}{N}, $$ where $$\sum\limits_{k=0}^{n} \dfrac{ \binom{K-1}{k} \binom{N-K}{n-k}}{\binom{N-1}{n}}=1$$ follows from Vandermonde's identity. So, $P_{2}=P_{50} = P_{100} = \dfrac{K}{N} = \dfrac{r}{100}$.