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.