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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 $P(X = 4)$?

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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.

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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.

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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.

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$P(X=n)$ 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.

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