probability last draw I have a bag initially containing $r$ red fruit pastilles (my favourites) and $b$ fruit pastilles of other colours. From time to time I shake the bag thoroughly and remove a pastille at random. (It may be assumed that all pastilles have an equal chance of being selected.) If the pastille is red I eat it but otherwise I replace it in the bag. After n such drawings, I find that I have only eaten one pastille. Show that the probability that I ate it on my last drawing is:
$$\frac{(r+b-1)^{n-1}}{(r+b)^n-(r+b-1)^n}$$
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My thoughts on this: Since each draw is independent from the others, the probability to draw a blue pastille is $\frac{b}{b+r}$, so the probability that the first $n-1$ draws are all blue is $(\frac{b}{b+r})^{n-1}$ and the probability to pick the red pastille at the last draw is $\frac{r}{b+r}$, making the final probability $(\frac{b}{b+r})^{n-1} \cdot \frac{r}{b+r}$, which is obviously not the sought probability.

I believe that an easier method will be by induction.
For n=1 the conditional probability must be 1 which it is if you substitute.
Assume that the formula holds true for n=k and prove that it is also true for n=k+1
Consider k pickings:
The tree will be k long and k high as follows:
rbbbbb........b (1st line)  $P_{d1} = \Big(\frac{b}{r+b}\Big)^{0}\times{r \over {r+b}} \times \Big(\frac{b}{r+b-1}\Big)^{k-1}$
brbbbb........b     $P_{d2} = \Big(\frac{b}{r+b}\Big)^{1}\times{r \over {r+b}} \times \Big(\frac{b}{r+b-1}\Big)^{k-2}$
bbrbbb........b     $P_{d3} = \Big(\frac{b}{r+b}\Big)^{2}\times{r \over {r+b}} \times \Big(\frac{b}{r+b-1}\Big)^{k-3}$
……….
bbbbbb.......rb (k-1 line) $P_{dk-1} = \Big(\frac{b}{r+b}\Big)^{k-2}\times{r \over {r+b}} \times \Big(\frac{b}{r+b-1}\Big)^{1}$
bbbbbb.......br (k line)    $P_{dk} = \Big(\frac{b}{r+b}\Big)^{k-1}\times{r \over {r+b}} $
The conditional probability for n=k is given by:
$P(k) =\frac{P_{dk}}{\sum_{i=1}^k P_{di}} =  \frac{(r+b-1)^{k-1}}{(r+b)^k-(r+b-1)^k}$   (Eq.1)
where $P_{dk}$ is the unconditional probability of the last row (k-1 bs followed by one red)
Now consider k+1 pickings. The tree will be exactly the same as above for the first k rows and k columns. The k+1 column will have k bs and 1 r while the k+1 row will also have k bs and 1 r.
The conditional probability P(k+1) will be given by:
$P(k+1) =\frac{P_{dk+1}}{\sum_{i=1}^{k+1} P_{di}}$           (Eq.2)
$P_{dk+1}=\big(\frac{b}{(r+b)^k} \times \frac{r}{(r+b)}\big)=\frac{b}{(r+b)} P_{dk}$  (Eq.3)
By studying the two trees we can establish a simple relationship between the sum of unconditional probabilities of the (k+1)x(k+1) tree to that of the kxk tree. To go from the kxk to the (k+1)x(k+1) tree we simply have to multiply each row of the kxk tree by the probability for the added b in the k+1 column $\big(\frac{b}{(r+b−1)}\big)$ and then add the last row from the (k+1)x(k+1) tree. ie:
$\sum_{i=1}^{k+1} P_{di} = \frac{b}{(r+b-1)} \times \sum_{i=1}^{k} P_{di} + P_{dk+1}$
Substituting from Eq.1:
$\sum_{i=1}^{k+1} P_{di} = \frac{b}{(r+b-1)} \times \frac{P_{dk}}{P(k)} + P_{dk+1}$
Substituting from Eq.3:
$ \sum_{i=1}^{k+1} P_{di} = \frac{b}{(r+b-1)} \times \frac{P_{dk+1}}{P(k)} \frac{(r+b)}{b} + P_{dk+1}$
Therefore:
$\sum_{i=1}^{k+1} P_{di} = P_{dk+1} \big( \frac{r+b}{(r+b-1) P(k)} +1 \big)$
From Eq.2:
$\frac{1}{P(k+1)} = \frac{\sum_{i=1}^{k+1}P_{di}}{P_{dk+1}} = \frac{r+b}{(r+b-1)}\frac{1}{P(k)}+1$
Substituting P(k) from Eq.1:
$\frac{1}{P(k+1)} = \frac{(r+b)\big((r+b)^k-(r+b-1)^k\big)}{(r+b-1)(r+b-1)^{k-1}}+1 = $$\frac{(r+b)^{k+1}-(r+b)(r+b-1)^k+(r+b-1)^k}{(r+b-1)^k}=\frac{(r+b)^{k+1}-(r+b-1)(r+b-1)^k}{(r+b-1)^k} =$ $\frac{(r+b)^{k+1}-(r+b-1)^{k+1}}{(r+b-1)^k}$
Therefore: $P(k+1) = \frac{(r+b-1)^k}{(r+b)^{k+1}-(r+b-1)^{k+1}}$
QED
