A box contain $n$ coins Let $P(E_{i})$ be the probability that $i$ out of $n$ coins are biased. If $P(E_{i})$ is A box contain $n$ coins . Let $P(E_{i})$ be the probability that $i$ out of $n$ coins are biased. If $P(E_{i})$ is 
Directly proportional to $i(i+1)\;,  1\leq i\leq n$
$(a)$ Let $P$ be the probability that a coin selected at random is biased, Then $\lim_{n\rightarrow \infty}P$ is 
$(b)$ If a coin selected at random is found to be biased Then the probability that it is the only biased 
coin in the box is
$\bf{Attempt:}$ Given $\displaystyle P(E_{i}) = \frac{i}{n}$ and let  $P(E_{i}) = ki(i+1)$
so $$\sum^{n}_{i=1}P(E_{i}) = k\sum^{n}_{i=1}i(i+1) = k\bigg[\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}\bigg]$$
so $$1=k\frac{n(n+1)}{2}\bigg[\frac{2n+1}{3}+1\bigg] = \frac{k}{3}\cdot (n+1)(n+2)$$
so $\displaystyle k = \frac{3}{(n+1)(n+2)}$
Could some help  me how to solve it, Thanks
 A: $P$ is the probabity that you select one from the biased coins among the collection of $n$, where the count of biased coins is itself a random variable. 
That is $P=\mathsf P(B)$ where $B$ represents the event that the selected coin is biased.  
$E_i$ is the event that $i$ equals the count of biased coins among the $n$.   When given the count of biased coins we know the conditional probability for selecting a biased coin:
$$\mathsf P(B\mid E_i)=i/n$$
We were told, for all integer $i: 1\leq i\leq n$, that $\mathsf P(E_i) = k i(i+1)$ for some $k$, and you have determined that since $\sum\nolimits_{i=1}^n ki(i+1)=1$, then: $$k = \frac 3{n(n+1)(n+2)}$$
So , as by the Law of Total Probability, $P = \sum\nolimits_{i=1}^n \mathsf P(B\mid E_i)~\mathsf P(E_i)$, we have: $$P~{= \frac{3\sum\nolimits_{i=1}^ni^2(i+1)}{n^2(n+1)(n+2)} \\= \frac{3n+1}{4n}}$$
Now $\lim\limits_{n\to\infty} P$ should be obvious.

Now find $\mathsf P(E_1\mid B)$, the probability for the count of biased coins being 1 when given that we have selected a biased coin.
