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.
 A: To get the needed probability, you need to divide the probability of getting exactly one on the last pick by the probability of getting exactly one in all the picks.
The probability of drawing one and only one red on the first of n picks is $${\frac{r}{r+b}} \times \Big(\frac{b}{r+b-1}\Big)^{n-1}$$
The probability of drawing the red on the second pick is $${\frac{b}{r+b}}\times{\frac{r}{r+b}} \times \Big(\frac{b}{r+b-1}\Big)^{n-2}$$
In general, then, the probability of drawing the one and only red on the i-th pick is $$P_n(i) = \Big(\frac{b}{r+b}\Big)^{i-1}\times{r \over {r+b}} \times \Big(\frac{b}{r+b-1}\Big)^{n-i}$$
So, the probability of drawing exactly one red in n tries is
$$\sum_{i=1}^n P_n(i)$$
and the probabiliy of getting that one red on the last pick is
$$\frac{P_n(n)}{\sum_{i=1}^n P_n(i)}$$
Grinding though the expanded summation then simplifying the result is a royal pain, but it does in fact reduce to the expression provided in the question.
A: 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
