Biased coins!

We're given that $P(E_k)$ is proportional to $k(k+1).$ Then $P(E_k)=ak(k+1)$ for some $a.$ We can also be sure that the sum of all $P(E_k)$ is equal to $1.$ Then,

$\begin{aligned} 1 &= \sum\limits_{k=0}^n{ak(k+1)} \\ &= a\sum\limits_{k=1}^n{(k^2+k)} \\ &= a\left[ \frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2} \right] \\ &= a\left[ \frac{n(n+1)(n+2)}{3} \right] \\ \end{aligned}$

This gives $a=\dfrac{3}{n(n+1)(n+2)}.$ Then,

$P(E_k)=\frac{3k(k+1)}{n(n+1)(n+2)}$

Let $B$ be the event that a biased coin is picked from the $n$ coins. This event is dependent on how many biased coins there are.

$P(B \mid E_k)=\frac{k}{n}$

$P(B \cap E_k)=\frac{k}{n} \cdot P(E_k)$

These probabilities are mutually exclusive across all $k,$ so $P(B)$ can be computed as the sum of all these probabilities.

$\begin{aligned} P(B) &= \sum\limits_{k=0}^n{P(B \cap E_k)} \\ &= \sum\limits_{k=0}^n{\frac{k}{n} \cdot \frac{3k(k+1)}{n(n+1)(n+2)}} \\ &= \frac{3}{n^2(n+1)(n+2)} \sum\limits_{k=1}^n{(k^3+k^2)} \\ &= \frac{3}{n^2(n+1)(n+2)} \left[\frac{n^2(n+1)^2}{4}+\frac{n(n+1)(n+2)}{6} \right] \\ &= \frac{3n+1}{4n} \end{aligned}$

Now we have everything we need to compute the desired probability. We'd like the probability that we pull a biased coin, given that there is only 1 biased coin. In notation, this is $P(B \mid E_1).$

$\begin{aligned} P(B \mid E_1) &= \frac{P(B \cap E_1)}{P(B)} \\ &= \frac{6}{n^2(n+1)(n+2)} \cdot \frac{4n}{3n+1} \\ &= \boxed{\dfrac{24}{n(n+1)(n+2)(3n+1)}} \end{aligned}$