Negative binomial distribution

DotKite
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Homework Statement



Repeatdly roll a fair die until the outcome 3 has accurred on the 4th roll. Let
X be the number of times needed in order to achieve this goal. Find E(X)
and Var(X)

Homework Equations





The Attempt at a Solution



I am having trouble deciphering this question. Is the first sentence saying to roll a die until you get a 3 on the fourth roll? thus the event of a success is when you get a 3 on the fourth roll? but what is number of successes? which you need to know in order to find the mean in a negative binomial distribution.

Apparently E(X) = 24.
 
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DotKite said:

Homework Statement



Repeatdly roll a fair die until the outcome 3 has accurred on the 4th roll. Let
X be the number of times needed in order to achieve this goal. Find E(X)
and Var(X)

Homework Equations





The Attempt at a Solution



I am having trouble deciphering this question. Is the first sentence saying to roll a die until you get a 3 on the fourth roll? thus the event of a success is when you get a 3 on the fourth roll? but what is number of successes? which you need to know in order to find the mean in a negative binomial distribution.

Apparently E(X) = 24.

Please clarify: do you mean the first 1, 2 or 3 tosses could also result in a "3", and that you look only at the result of the fourth toss? Or, do you mean that you count the rolls only if the first three are non-3s and the fourth is a 3? It makes a huge difference!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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