Probability - Poisson Random Variable

rooski
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Homework Statement



A trial consists of tossing two dice. The result is counted as successful if the sum of the
outcomes is 12. What is the probability that the number of successes in 36 such trials is
greater than one? What is this probability if we approximate its value using the Poisson
random variable?

The Attempt at a Solution



So first we must calculate the probability that the dice will add up to 12. The only possible way for this to happen is if they are both showing a 6.

So multiplying the chances of a 6 showing up per dice we get 1/6 * 1/6, or 1/36. There is a 1/36 chance of seeing 12.

This is where i get confused. How do we calculate the probability that 36 trials will give us more than 1 event where the dices read 12? I'm also confused about the Poisson Random Variable part.
 
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The binomial distribution gives you the probabilities you're seeking. Start there.

In some cases, the binomial distribution is well approximated by the Poisson distribution, with λ=np. The second part of the problem wants you to calculate the same probability you did before using this approximation.
 
So the number of trials, n, is 36 and the probability of success, p, is 1/36, right?
 
Yup.
 
Ok i figured out the first part of the question - the probability of the trials having more than 1 success I am 36 trials is 0.2642 using the binomial distribution function.

How do i apply the poisson variable to this?
 
Look up how to approximate the binomial distribution using the Poisson distribution in your textbook.
 
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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