Calculating Lottery Probability for Drawing a Specific Number

[Gregg]

Homework Statement

In a random number game, the lottery, I need to know the probability of a number being drawn over a number of draws.

Each ball of has a chance of 1 in 7 of being drawn in a draw. If the expected frequency of each ball is 200 (over 1400 draws) I need to know how to work out the probability of a ball being drawn 170 or 230 times rather than the expected 200.

The Attempt at a Solution

All I can think of is using the normal distribution but I can't see how to use it here. I have no s.d. also this is a discrete random variable.

 

[phyzguy]

What you want is called a Poisson distribution, You can look it up on Wikipedia. It is:
P(N0,N)=\frac{N0^N e^{-N0}}{N!}
Where N0 is the expected number, and N is the number you want the probability of. For the numbers you gave, both probablilities are ~0.3%. Remember this is just the probability of getting exactly 170. If you want the probability of getting <170, then you need to add up all of the probabilities for N<=170.

 

[Gregg]

Why is it Poisson distribution as opposed to other distributions?

 

[phyzguy]

The Poisson distribution governs probabilities of events which are discretely distributed - meaning here that a number can be drawn 200 times or 201 times, but not 200.3748 times - and assumes that the events are independent of one another. Continuous variables (for example the distribution of height of a large number of people) tend to follow Gaussian distributions. As to why - Wikipedia has a pretty good write-up on how the Poisson distribution arises.

 

[Gregg]

So the probability of a ball being drawn less than the mean will be 0.5. For 49 balls the probability will be greater than 1?

 

[vela]

You know the probability of a ball being drawn in a trial and the number of trials, so you should be looking at the binomial distribution.

 

[Gregg]

So the probability at first is found with Poisson, then you use binomial with that probability?

 

[vela]

No. You don't need the Poisson distribution at all. (I don't think it really applies to this problem anyway.)

 

[phyzguy]

Actually, in this case, if you know the number of trials, I think Vela is right, the binomial distribution is more appropriate. The Poisson distribution is the limit of the binomial distribution as the number of trials gets large. For the numbers in your original question, the difference between the two is pretty small.

 

[Gregg]

OK well the probability of a ball being drawn 177 times out of 1494.

\left(\frac{1}{7}\right)^x \left(\frac{6}{7}\right)^{1494-x} \left(<br /> \begin{array}{c}<br /> 1494 \\<br /> x<br /> \end{array}<br /> \right)=0.000702881
with x=177.

If there are 49 balls then the probability is 49(0.000702881) isn't it?

To find the probability of the frequency of a ball being 213 or less we have

\sum _{x=0}^{213} \left(\frac{1}{7}\right)^x \left(\frac{6}{7}\right)^{1494-x} \left(<br /> \begin{array}{c}<br /> 1494 \\<br /> x<br /> \end{array}<br /> \right) = 0.5

If I want to find the probability that any of the 49 balls a drawn 213 or less times, then 49*0.5 is larger than 1?

So is the first part wrong?

 

[vela]

OK well the probability of a ball being drawn 177 times out of 1494.

\left(\frac{1}{7}\right)^x \left(\frac{6}{7}\right)^{1494-x} \left(<br /> \begin{array}{c}<br /> 1494 \\<br /> x<br /> \end{array}<br /> \right)=0.000702881
with x=177.

If there are 49 balls then the probability is 49(0.000702881) isn't it?

No. The events aren't disjoint. For example, just because ball 1 appears 177 times doesn't mean that ball 2 doesn't also appear 177 times. You can only sum probabilities of events if the events are disjoint.

To find the probability of the frequency of a ball being 213 or less we have

\sum _{x=0}^{213} \left(\frac{1}{7}\right)^x \left(\frac{6}{7}\right)^{1494-x} \left(<br /> \begin{array}{c}<br /> 1494 \\<br /> x<br /> \end{array}<br /> \right) = 0.5

If I want to find the probability that any of the 49 balls a drawn 213 or less times, then 49*0.5 is larger than 1?

So is the first part wrong?

 

[Gregg]

So how do I find the probability that any of the 49 balls will be drawn less than a certain amount of times?

 

[Gregg]

\sum _{n=0}^{177} \left(\frac{1}{7}\right)^n\left(\frac{6}{7}\right)^{1494-n}\left(<br /> \begin{array}{c}<br /> 1494 \\<br /> n<br /> \end{array}<br /> \right)=0.00331939

<br /> (0.00331939)^1(1-0.00331939)^{49-1}\left(<br /> \begin{array}{c}<br /> 49 \\<br /> 1<br /> \end{array}<br /> \right)=0.1386573

Is this correct?

 

[Gregg]

I don't think it is, still no answer.