# Homework Help: Lottery probability question

1. Mar 30, 2010

### Gregg

1. The problem statement, all variables and given/known data

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.

3. 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.

2. Mar 30, 2010

### 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.

3. Apr 4, 2010

### Gregg

Why is it Poisson distribution as opposed to other distributions?

4. Apr 4, 2010

### 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.

5. Apr 5, 2010

### 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?

6. Apr 5, 2010

### vela

Staff Emeritus
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.

7. Apr 5, 2010

### Gregg

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

8. Apr 5, 2010

### vela

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

9. Apr 5, 2010

### 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.

10. Apr 5, 2010

### Gregg

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

$$\left(\frac{1}{7}\right)^x \left(\frac{6}{7}\right)^{1494-x} \left( \begin{array}{c} 1494 \\ x \end{array} \right)=0.000702881$$
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

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

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?

11. Apr 5, 2010

### vela

Staff Emeritus
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.

12. Apr 5, 2010

### Gregg

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

13. Apr 6, 2010

### Gregg

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

$$(0.00331939)^1(1-0.00331939)^{49-1}\left( \begin{array}{c} 49 \\ 1 \end{array} \right)=0.1386573$$

Is this correct?

14. Apr 6, 2010

### Gregg

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