91Ó°ÊÓ

Beginning from $a$white spheres plus $b$black spheres, $E_{a,b}$- the white sphere was that final one inside the container.

Every white spheres were selected initially $W$.

Every black spheres were selected initially $B$.

role="math" localid="1649435638601" $a,b∈\mathrm{ℕ}=\{1,2,3,…\}$

Base: to any or all $a,b∈\mathrm{ℕ}$so $a+b=n=1$, which is that $a=b=1$equation is correct, the balls inside urn was white by probability $\frac{1}{2}$

Assume that $n∈\mathrm{ℕ}$and to any or all $a,b∈\mathrm{ℕ}$so $a+b<n$.

role="math" localid="1649436772793" ${\mathrm{P}}_{\mathrm{a},\mathrm{b}}=\frac{1}{2}$

Consider $a,b∈\mathrm{ℕ}$so $a+b=n:$

Because $W,B,W^c{B}^c$were completely exclusive occurrences which which incorporates the whole outcome area, we may use the entire probabilistic equation:

$P\left(E_{a,b}\right)=P\left(E_{a,b}âˆ\pounds W^c{B}^c\right)P\left(W^c{B}^c\right)+P\left(E_{a,b}âˆ\pounds B\right)P\left(B\right)+P\left(E_{a,b}âˆ\pounds W\right)P\left(W\right)$

If although neither a black nor white spheres were eliminated. Especially at starting, will have at minimum a ball less, and method to get rid of the spheres begins.

The probability are easy when all spheres of a colour were taken:

$P\left(E_{a,b}âˆ\pounds W\right)=0P\left(E_{a,b}âˆ\pounds B\right)=1$

localid="1649628952164" $\begin{array}{c}\mathrm{P}\left(\mathrm{W}\right)=\frac{\mathrm{a}}{\mathrm{a}+\mathrm{b}}×\frac{\mathrm{a}−1}{\mathrm{a}+\mathrm{b}−1}â‹\dots ⋯\frac{1}{\mathrm{b}+1}=\frac{\mathrm{a}!}{\frac{(\mathrm{a}+\mathrm{b})!}{\mathrm{b}!}}=\frac{\mathrm{a}!\mathrm{b}!}{(\mathrm{a}+\mathrm{b})!}\\ \mathrm{P}\left(\mathrm{B}\right)=\frac{\mathrm{b}}{\mathrm{a}+\mathrm{b}}×\frac{\mathrm{b}−1}{\mathrm{a}+\mathrm{b}−1}â‹\dots ⋯\frac{1}{\mathrm{a}+1}=\frac{\mathrm{b}!}{\frac{(\mathrm{a}+\mathrm{b})!}{\mathrm{a}!}}=\frac{\mathrm{a}!\mathrm{b}!}{(\mathrm{a}+\mathrm{b})!}\\ \mathrm{P}\left(\mathrm{W}\right)=\mathrm{P}\left(\mathrm{B}\right)\end{array}$

So,

$\mathrm{P}\left({\mathrm{E}}_{\mathrm{a},\mathrm{b}}\right)=\mathrm{P}\left({\mathrm{E}}_{\mathrm{a},\mathrm{b}}âˆ\pounds {\mathrm{W}}^{\mathrm{c}}{\mathrm{B}}^{\mathrm{c}}\right)\mathrm{P}\left({\mathrm{W}}^{\mathrm{c}}{\mathrm{B}}^{\mathrm{c}}\right)+\mathrm{P}\left({\mathrm{E}}_{\mathrm{a},\mathrm{b}}âˆ\pounds \mathrm{B}\right)\mathrm{P}\left(\mathrm{B}\right)+\mathrm{P}\left({\mathrm{E}}_{\mathrm{a},\mathrm{b}}âˆ\pounds \mathrm{W}\right)\mathrm{P}\left(\mathrm{W}\right)$

localid="1649439452847" $\begin{array}{r}=\frac{1}{2}\mathrm{P}\left({\mathrm{W}}^{\mathrm{c}}{\mathrm{B}}^{\mathrm{c}}\right)+0+1×\mathrm{P}\left(\mathrm{W}\right)\\ =\frac{1}{2}\mathrm{P}\left({\mathrm{W}}^{\mathrm{c}}{\mathrm{B}}^{\mathrm{c}}\right)+\frac{1}{2}×2\mathrm{P}\left(\mathrm{W}\right)\end{array}$

localid="1649439475979" $\begin{array}{r}=\frac{1}{2}\mathrm{P}\left({\mathrm{W}}^{\mathrm{c}}{\mathrm{B}}^{\mathrm{c}}\right)+\frac{1}{2}×\left(\mathrm{P}\right(\mathrm{W})+\mathrm{P}(\mathrm{B}\left)\right)\\ \end{array}$

$=\frac{1}{2}\left(\mathrm{P}\left({\mathrm{W}}^{\mathrm{c}}{\mathrm{B}}^{\mathrm{c}}\right)+\mathrm{P}\left(\mathrm{W}\right)+\mathrm{P}\left(\mathrm{B}\right)\right)$

$=\frac{1}{2}×1$

$=\frac{1}{2}$