It is known that an urn containing altogether 10 balls was filled in the following manner: A coin was tossed 10 times, and according as it showed heads or tails, one white or on black ball was put into the urn. Balls are drawn from this urn one at a time, 10 times in succession (with replacements) and every one turns out to be white. What is the chance that the urn contains nothing but white balls?
My work:
Let $A_i$ denote the event that the urn contains $i$ white balls, where $i$ assumes values from 0 to 10.
In $10$ throws, if the coin show $i$ heads and $10-i$ tails, then $$P(A_i)= {}^{10}\!C_i(1/2)^i (1/2)^{10−i} = {}^{10}\!C_i (1/2)$$
Let $A$ denote the event of drawing a white ball, then.
$P(A∣B_i)=i/10$ where $i=0,1,2,...10$
Hence , using Baye's formula
$P(A_{10}∣A)= (1/2)^9$
However, the answer given is $0.0702$. Where I am wrong.
It is hard to say where you are wrong because it is difficult to distinguish between misprints and errors in your text. For example you use a variable $B_i$, which was never defined, and use the symbol $A$ both for the event of having some amount of the balls in the urn and for the event to draw some amount of balls from the urn.
Let $A_n$ be event that the urn contains $n$ white balls, the probability of which is $$ \mathsf P(A_n)=\binom {10}{n}\frac1 {2^{10}} . $$
In this case the probability of the event $B$ of drawing 10 times a white ball is $$\mathsf P(B|A_n)=\left (\frac n {10}\right)^{10},$$ and the overall probability $$ \mathsf P(B)=\sum_{n=0}^{10}\mathsf P(B|A_n). $$
Then by the Bayes formula the probability that the urn contains only white balls is $$ \mathsf P(A_{10}|B)=\frac{\mathsf P(B|A_{10})\mathsf P(A_{10})}{\mathsf P(B)}= \frac {\left(\frac{10}{10}\right)^{10} \cdot \binom {10}{10}\frac1 {2^{10}}} {\frac1{2^{10}}\sum_{n=0}^{10}\binom {10}{n}\left (\frac n {10}\right)^{10} }=\frac {7812500}{111304237}. $$