Question: Let $\{a\_n\}$ be the sequence of Fibonacci numbers, i.e. $a\_1 = a\_2 = 1, a\_n = a\_{n-1} + a\_{n-2}, n \ge 3$. Find $$\lim\_{n\to\infty}\sqrt[n]{a\_n}.$$

Answer: 
(1) Each ball is drawn without replacement. 

Suppose we continue to draw until all balls are drawn. After the first ball is drawn, there are 19 remaining draws to extract the 4 balls of that same colour. Calculate the probability that no such are drawn before draw $X$, one on draw number $X$, and that *the remaining $3$ are drawn in the last $20-X$ draws*.

(2) Each ball is drawn with replacement. 

Calculate the probability that draws numbered $2$ to $X-1$ are each of a different colour to, and draw numbered $X$ is of the same colour as, the first ball drawn.
My answer to (2):

Suppose the first ball chosen was green, Now we need to choose two more green balls and we need to choose a non-green ball on the 4-th selection.

Now if we got a green ball on the first selection we can define a geometric random variable $Y$ with parameter $p=\frac{15}{20}=\frac{3}{4}$ (since we have $15$ non-green balls) 

And we get $\forall n\in\{1,2,3...\}, P\{Y=n\}=(1-p)^{n-1}p=(\frac{1}{4})^n\frac{3}{4}$.

Now since there are four ways to choose the first ball, we have to multiply that result by $4$.

And we get that $4\times P\{Y=3\}=4\times\frac{3}{64}=\frac{12}{64}$

And so $P\{X=4\}=4\times P\{Y=3\}=\frac{12}{64}$ which is the required probability.