91Ó°ÊÓ

$C_i$ - the coin with probability $\frac{i}{k}$ of flipping heads is chosen, $i=0,1,…,k$.

$F_{n,m}$ - the first $n+m$ flips resulted in $n$ heads and $m$ tails.

$H-n+m+1$. flip is heads.

Calculation of $P\left(Hâˆ\pounds F_{n,m}\right)$

Same logic as in Example 5e: Conditioning on which coin is chosen, we get the formula:

$P\left(Hâˆ\pounds F_{n,m}\right)=\underset{i=0}{\overset{k}{∑}}P\left(Hâˆ\pounds C_i{F}_{n,m}\right)P\left(C_iâˆ\pounds F_{n,m}\right)$

After applying formula for conditional probability and formula for total probability, as in the Example 5e:

$P\left(C_iâˆ\pounds F_{n,m}\right)=\frac{\left(\begin{array}{c}n+m\\ n\end{array}\right){\left(\frac{i}{k}\right)}^n{\left(1-\frac{i}{k}\right)}^m}{{∑}_{j=0}^k\left(\begin{array}{c}n+m\\ n\end{array}\right){\left(\frac{j}{k}\right)}^n{\left(1-\frac{j}{k}\right)}^m}$

Substituting this in formula (1) this is obtained:

$P\left(Hâˆ\pounds F_{n,m}\right)=\underset{i=0}{\overset{k}{∑}}\frac{\left(\begin{array}{c}n+m\\ n\end{array}\right){\left(\frac{i}{k}\right)}^{n+1}{\left(1-\frac{i}{k}\right)}^m}{{∑}_{j=0}^k\left(\begin{array}{c}n+m\\ n\end{array}\right){\left(\frac{j}{k}\right)}^n{\left(1-\frac{j}{k}\right)}^m}$

More nicely put:

Let $C_{n,m}$denote the integral approximation of the following expression:

$\frac{1}{k}\underset{i=0}{\overset{k}{∑}}{\left(\frac{i}{k}\right)}^n{\left(1-\frac{i}{k}\right)}^m≈{∫}_0^1{y}^n(1-y{)}^{m}dy=:C_{n,m}$

Then the wanted probability would be approximately:

$P\left(Hâˆ\pounds F_{n,m}\right)≈\frac{C_{n+1,m}}{C_{n,m}}$

As hinted, first use partial integration :

$={∫}_0^1\frac{1}{n+1}{u}^{\text{'}}\left(y\right)v\left(y\right)dy$

$=\frac{1}{n+1}\left[{\left.\left(u\right(y\left)v\right(y\left)\right)\right|}_0^1-{∫}_0^{1}u\left(y\right)v^{\text{'}}\left(y\right)dy\right]$

$=\frac{1}{n+1}\left[0-{∫}_0^1{y}^{n+1}·(-m)(1-y{)}^{m-1}dy\right]$

$=\frac{m}{n+1}{∫}_0^1{y}^{n+1}(1-y{)}^{m-1}dy$

$=\frac{m}{n+1}{C}_{n+1,m-1}$

And integration shows that:

$C_{n,0}={∫}_0^1{y}^{n}dy=\frac{1}{n+1}$

By repeating recursion $m$times, until $0$is reached as the second index the formula becomes explicit:

$C_{n,m}=\frac{m}{n+1}·\frac{m-1}{n+2}·\frac{m-2}{n+3}·⋯·\frac{1}{n+1+m-1}·C_{n+m,0}$

$=\frac{m!}{\frac{(n+m)!}{n!}}·\frac{1}{n+m+1}$

$=\frac{n!m!}{(n+m+1)!}$

Now returning to the wanted probabiltiy $P\left(Hâˆ\pounds F_{n,m}\right)$

The required probability is$\left.P\left(Hâˆ\pounds F_{n,m}\right)≈\frac{C_{n+1,m}}{C_{n,m}}\right)=\frac{n+1}{n+m+2}$.