# Thread: Possible Binomial Distribution Question

1. ## Possible Binomial Distribution Question

A standard, fair die is thrown repeatedly. Let $U$ denote the number of throws before a six occurs. Choose a standard probability distribution to model this and use it to answer the following.
a) What is the probability function of $U$.
b) Find the mean of $U$.
c) Find the variance of $U$.

a) I presume that this has a binomial distribution, $X = B(U,\frac{1}{6})$.

b) The mean would then be $U \times \frac{1}{6} = \frac{U}{6}$

c) From what I can remember the variance is $np(1-p)$, this would give $U \times \frac{1}{6}(1-\frac{1}{6}) = \frac{5U}{36}$

Is this correct? The only thing I'm a bit unsure about is the part of the question that says, "before a six occurs", not sure if you have to do something different because of this?

Thanks again

2. It is a geometric distribution that is based on the binomial distribution and a single Bernoulli trial. The binomial distribution computers the probability of getting X successes in N throws. The geometric distributions says "After X-1 throws, what is the probability that I will get a success on the Xth throw?" So it is a P(X-1 throws with no successes AND Success on the Xth throw).

Armed with this do you think you can figure out what the function would be?

3. Originally Posted by ANDS!
It is a geometric distribution that is based on the binomial distribution and a single Bernoulli trial. The binomial distribution computers the probability of getting X successes in N throws. The geometric distributions says "After X-1 throws, what is the probability that I will get a success on the Xth throw?" So it is a P(X-1 throws with no successes AND Success on the Xth throw).

Armed with this do you think you can figure out what the function would be?

So is it not the Binomial distribution then? Sorry it's been a long time since I've attempted any statistics. I get what you saying about the success on the Xth throw and how this fits the question.

Unsure where to go from here though?

Thanks again

4. It is sort of like the binomial distribution. Remember with the binomial that we are not just interested in say (in your case) QQQQQP, where Q is a failure, and P is a success, because I could have PQQQQQ, and QQPQQQ, where each arrangement of P's and Q's is the probability of getting one six in six throws. For a binomial, we are interested in ALL possible ways of getting one six in six throws. What the geometric says, is we are interested in that sequence where I have thrown the die 6 times, and I want the probability that the next throw is a success: This would be P(Five failures) and P(Success on Next try); since these two events are independent - we just multiply the two probabilities.

More generally we define a geometric random variable as: $X~Geo(p)$ where p is the success probability, and X is the number of throws it takes you to get a success. The probability mass function for the random variable X is then: $P(X=x) = p*q^{x-1}$. With a little digging you can see that this is a "binomial distribution" with "n choose k" of zero (where "n" is equal to X-1 from the geometric distribution): $\frac{n!}{0!(n-0)!}p^{0}q^{n-0}$, multiplied by "p".

Hopefully this makes some sense and helps to tie all of this together.

5. Originally Posted by ANDS!
It is sort of like the binomial distribution. Remember with the binomial that we are not just interested in say (in your case) QQQQQP, where Q is a failure, and P is a success, because I could have PQQQQQ, and QQPQQQ, where each arrangement of P's and Q's is the probability of getting one six in six throws. For a binomial, we are interested in ALL possible ways of getting one six in six throws. What the geometric says, is we are interested in that sequence where I have thrown the die 6 times, and I want the probability that the next throw is a success: This would be P(Five failures) and P(Success on Next try); since these two events are independent - we just multiply the two probabilities.

More generally we define a geometric random variable as: $X~Geo(p)$ where p is the success probability, and X is the number of throws it takes you to get a success. The probability mass function for the random variable X is then: $P(X=x) = p*q^{x-1}$. With a little digging you can see that this is a "binomial distribution" with "n choose k" of zero (where "n" is equal to X-1 from the geometric distribution): $\frac{n!}{0!(n-0)!}p^{0}q^{n-0}$, multiplied by "p".

Hopefully this makes some sense and helps to tie all of this together.
Thanks again for yours answer, just been to find some notes on Geometric distributions.

As you said earlier, the answer is a geometric distribution, $X ~ Ge(\frac{1}{6})$

I have in my notes that $E[X] = \frac{1}{p}$ and $Var[X] = \frac{1-p}{p^2}$, so using these would my mean be $6$ and variance be $\frac{1-\frac{1}{6}}{\frac{1}{6}^2} = 30$?

Thanks again for the help