Question

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. The chance that the urn contains nothing but white balls is $\frac{1}{2^k}$. Find the value of $k$.

Answer 1

Let $B_k$ denote the event that the urn contains $k$ white balls, where $k$ assumes values from $0$ to $10$.

In $10$ throws, if the coin show $k$ heads and $10-k$ tails, then
$P\left(B_k\right){=}^{10}{C}_k{\left(\frac{1}{2}\right)}^k{\left(\frac{1}{2}\right)}^{10-k}{=}^{10}{C}_k\frac{1}{2^{10}}$
Let $A$ denote the event of drawing a white ball, then.
$P(A\mid B_k)=\frac{k}{10}$ where $k=0,1,2,...,10$
Hence using Baye's formula
$p(B_{10}\mid A)=\frac{P\left(B_{10}\right)P(A\mid B_{10})}{\sum _{k=0}^{10}P\left(B_k\right)P(A\mid B_k)}$
$=\frac{\left({}^{10}{C}_{10}\frac{1}{2^{10}}\right)\frac{10}{10}}{\sum _{k=0}^{10}{}^{10}{C}_k\frac{1}{2^{10}}\frac{k}{10}}=\frac{1}{2^9}$

applying the formula in the denominator $n{.2}^{n-1}=\sum _{k=1}^{n}k{.}^n{C}_k$ of binomial theorem with $n=10.$