Consider a discrete-time counting process $X_n$ with $n = 0,1,...$ so that $X_0 = 0$ and $X_n = X_{n-1} + \xi_n$ for each $n \geq 1$, where $\xi_1,\xi_2,\ldots$ are iid Bernoulli random variables with the common probability of success $p = 0.4$. Given that $X_8 = 2$ compute: $P(X_{14} = 4 | X_8 = 2)$.
So far I have tried that since we already have 2 successes at 8 trials, this means that we need 2 more in the trials $n = 9$ to $n = 14$ and so this would be with probability $0.4*0.4*0.6*0.6*0.6*0.6$ where there are 2 successes and 4 fails which would be enough since we are trying to get 4 successes in total at $n=14$. However, this was marked as incorrect. I would appreciate if someone could lead me in the right direction.