A popular problem by Brilliant Member

Let $p$ be the probability that a problem posted by Billy Bob Joe will become popular. It follows that $1 - p$ is the probability that a problem posted by Billy Bob Joe will not become popular.

The probability that exactly 2 of Billy Bob Joe's $n$ problems will become popular is equal to $\displaystyle \binom{n}{2} p^2 (1 - p)^{n - 2},$ and the probability that exactly 3 of the problems will become popular is equal to $\displaystyle \binom{n}{3} p^3 (1 - p)^{n - 3}.$ The binomial coefficients are there because we must select which three (or which two) problems will become popular.

From the problem statement, we have that

$\begin{aligned} n \binom{n}{2} p^2 (1 - p)^{n - 2} &= \binom{n}{3} p^3 (1 - p)^{n - 3} \\ n \binom{n}{2} (1 - p) &= \binom{n}{3} p \\ n \left ( \dfrac{n!}{2!(n - 2)!} \right ) (1 - p) &= \dfrac{n!}{3!(n - 3)!} p \\ n(1 - p) &= \dfrac{n - 2}{3} p \\ 3n - 3np &= (n - 2)p \\ 3n &= (4n - 2)p \\ p &= \dfrac{3n}{4n - 2}. \end{aligned}$

As $n$ gets larger and larger, the value of $p$ will approach $\displaystyle \lim_{n \rightarrow \infty} \dfrac{3n}{4n - 2} =\dfrac{3}{4}.$ Thus, $a + b = 3 + 4 = \boxed{7}.$