Question

A bag contains $n$ balls; $k$ drawings are made in succession, and the ball on each occasion is found to be white: find the chance that the next drawing will give a white ball; (i) when the balls are replaced after each drawing; (ii) when they are not replaced.

Answer 1

(i) Before the observed event there are $n+1$ hypotheses, equally likely; for the bag may contain $0,1,2,3,.....n$ white balls.
$P_0=P_1=P_2=P_3=.....=P_n$;
and $p_0=0,p_1={\left(\frac{1}{n}\right)}^k,p_2={\left(\frac{2}{n}\right)}^k,p_3={\left(\frac{3}{n}\right)}^k,........,p_n={\left(\frac{n}{n}\right)}^k$.
Hence after the observed event,
$Q_r=\frac{r^k}{1^k+2^k+3^k+.....+n^k}$
Now the chance that the next drawing will give a white ball $=\sum \frac{r}{n}{Q}_r$;
Thus the required chance $=\frac{1}{n}\cdot \frac{1^{k+1}+2^{k+1}+3^{k+1}+......+n^{k+1}}{1^k+2^k+3^k+.....+n^k}$;
and the value of numerator and denominator may be found.
In the particular case when $k=2$,
the required chance is $\frac{1}{n}{\left\{\frac{n(n+1)}{2}\right\}}^2÷\frac{n(n+1)(2n+1)}{6}$
$=\frac{3(n+1)}{2(2n+1)}$.
If $n$ is indefinitely large, the chance is equal to the limit, when $n$ is infinite, of $\frac{1}{n}\cdot \frac{n^{k+2}}{k+2}÷\frac{n^{k+1}}{k+1};$
and thus the chance is $\frac{k+1}{k+2}$.
(ii) If the balls are not replaced,
$p_r=\frac{r}{n}\cdot \frac{r-1}{n-1}\cdot \frac{r-2}{n-2}.........\frac{r-k+1}{n-k+1};$
and $Q_r=\frac{p_r}{\sum p_r}=\frac{(r-k+1)(r-k+2).....(r-1)r}{\sum _{r=0}^{r=n}(r-k+1)(r-k+2)........(r-1)r}$
$=(k+1)\frac{(r-k+1)(r-k+2).....(r-1)r}{(n-k+1)(n-k+2)..........(n-1)n(n+1)}.$
The chance that the next drawing will give a white ball $=\sum _{r=0}^{r=n}\frac{r-k}{n-k}{Q}_r$
$=\frac{k+1}{(n-k)(n-k+1).....n(n+1)}\sum _{r=0}^{r=n}(r-k)(r-k+1)..........(r-1)r$
$=\frac{k+1}{(n-k)(n-k+1).....n(n+1)}\cdot \frac{(n-k)(n-k+1).....n(n+1)}{k+2}$
$=\frac{k+1}{k+2}$,
which is independent of the number of balls in the bag at first.