6 sided die probabilities

i am currently working on a study guide and one of the questions i am completely stuck on and have no idea how to do it. Question is. You are interested in the number of rolls of a fair $6$ sided die until a number $2$ shows up.

Let $X =$ The number of times you roll the die until a number $2$ shows up.

(a) What type of random variable is $X$?

(b) How many rolls do you expect it to take? That is, what is the expected value, or mean, of the random variable $X$?

(c) What is the probability you roll a $2$ for the first time on the fourth roll? i.e. What is $P(X = 4)$?

Think of rolling the die as a Bernoulli trial, and a 2 as a success. The probability distribution that tells you how many trials until you get a success is a geometric distribution. The facts about this distribution will let you answer the rest of your questions.

Try to answer part (a). Once you have identified the distribution, the rest is fairly straightforward.

You are looking for the distribution of a count of tries until a success.

Some hints:

a) Random variables belong in two types: discrete and continuous.

b) This is a geometric distribution.

c) Two independent events must occur for this. You must not get a 2 in your first 3 trials. You must then get a 2 in your 4th trial. Each trial is independent.

$P(X=n)$ is $n-1$ rolls without seeing a $2$ followed by a $2.$

$P(X=n) = (\frac 56)^{n-1} \frac 16$

$E[X] = \sum_\limits{n=1}^{\infty} nP(X=n)$ which is a realative of the sum of an infinite geometric series.

That is $S = \sum_\limits{n=1}^{\infty} n r^n$

$rS = \sum_\limits{n=1}^{\infty} n r^{n+1} = \sum_\limits{n=1}^{\infty} (n-1) r^n\\ S - rS = \sum_\limits{n=1}^{\infty} n r^{n+1} = \sum_\limits{n=1}^{\infty} n r^n - (n-1) r^n = \sum_\limits{n=1}^{\infty} r^{n}$

Which is an ordinary geometric series.

$(1 - r)S = \frac r{1-r}\\ S = \frac r{(1-r)^2}$

Back to our case.

$E[X] = \sum_\limits{n=1}^{\infty} n \frac 16 (\frac56)^{n-1} = \frac 16 \frac 1{(1-\frac 56)^2} = 6$

Which is exactly what you thought it would be, before doing all that math.

What is the name for this distribution? I guess I would call it some variation of the beta distribution.