# Homework Help: Probability Distribution

1. Feb 20, 2012

### lina29

1. The problem statement, all variables and given/known data
Assume that the probability of error-free transmission of a message over a communication channel is p=0.9. If a message is not transmitted correctly, a retransmission is initiated. This procedure is repeated until a correct transmission occurs. Such a channel is often called a feedback channel. Assuming that successive transmissions are independent,

A- What is the probability that no retransmissions are required? (Note: Number of retransmissions = Number of transmissions - 1.)
B- What is the probability that exactly two retransmissions are required?
C- What is the expected number of retransmissions?
D- What is the variance of the number of retransmissions?

2. Relevant equations
A-.9
C-.1

3. The attempt at a solution
B- (1-p)(1-p)= (.1)(.1)=.01 which was wrong
D- Std(X)= sqrt( np (1-p)) = .1 and the variance is the std squared so var(X)=.01 which was also wrong

What did I miss?

2. Feb 20, 2012

### Dick

Well, for B you need to send the message three times. The first two have to fail and the third one has to succeed.

3. Feb 20, 2012

### lina29

so (.1)(.1)(.9)=.009?
What would I do for D?

4. Feb 20, 2012

### Dick

Where did you get your formula for Std? What's n? I don't think your mean is correct either. How did you get that?

5. Feb 20, 2012

### lina29

I got the formula from my notes for the binomial distribution. However, if it's wrong I can use another formula. I don't know n but the expected value is .1 ( which was right) and it is also equal is np. Is there another approach I should take?

6. Feb 20, 2012

### Dick

I think this is actually called a Negative Binomial Distribution. It's like a binomial distribution but the only events that are allowed are a series of failures followed by a success. Anything in your notes on that?

7. Feb 20, 2012

### lina29

nope. How would I do that?

8. Feb 20, 2012

### Dick

Either look it up or try and work it out from scratch. For example, I get that the expected number of retransmissions is 0*(.9)+1*(.1)*.9+2*(.1)^2*(.9)+3*(.1)^3*(.9)+...

9. Feb 20, 2012

### lina29

I looked it up and it said I would need to know the number of trials and the number of success. How did you find that? Also this would give the expected number. How would I go from there to the variance?

10. Feb 20, 2012

### Dick

If n is the number of retransmissions and p(n) is the probability of n retransmissions you find the expected number by summing n*p(n). You did p(2) in exercise B. You should be able to find that case in the sum. You would find the variance by summing n^2*p(n) and then subtracting the expected number squared. If this is a formula based course and they haven't taught you how to sum things like this, then I think you are missing some formulas.

11. Feb 20, 2012

### lina29

This isn't a formula based course so we only learn basic formulas. Since I got the expected number previously as .1 I got n=1/9 and then from there I found the variance to be .001111. Would that be right?

12. Feb 20, 2012

### Dick

I think there is something messed up with this question. I get the expected number to be 1/9=0.11111.... There's something wrong if it's saying .1 is correct. If you want to look at some formulas go to http://en.wikipedia.org/wiki/Negative_binomial_distribution
There are formulas for the mean and variance on the right. You want to use r=1 (since only one success is required) and p=.1. What do you get?

13. Feb 20, 2012

### lina29

it might be saying .1 is correct since 1/9 is .1 repeating. I got the variance to be .123456 using the formula pr/ ((1-p)^2)

14. Feb 20, 2012

### Dick

Well, I got 10/81. But that's .123456 to six decimal places. So yes, it's kind of funny those formulae weren't in your notes.

15. Feb 20, 2012

### lina29

thank you so much!

16. Feb 20, 2012

### Ray Vickson

Your answer is misleading: the distribution is the Geometric, not the negative binomial (although, of course, the geometric is a special case of the negative binomial!). Geometric = number of trials until the first success (variant: number before the first success). Negative binomial = sum of independent, identically-distributed geometrics. It is called the negative binomial because its distribution involves binomial coefficients with appropriate negative arguments.

RGV

17. Feb 20, 2012

### Dick

Ok, I give up. I'm not an expert at this but isn't the negative binomial a distribution where you count failures with constant probability until a fixed number of successes (or vice versa). I just summed the series by hand and checked they matched the numbers quoted for the negative binomial to the best of my understanding. I didn't need to invoke any negative numbers in the binomial coefficients because I didn't do it that way. What do you think is wrong? Doesn't geometric just tell you about the number of successes versus the number of failures without any specific stopping condition except the total number of trials? If not, I'll stop trying to answer questions like this.

Last edited: Feb 20, 2012
18. Feb 21, 2012

### Ray Vickson

The geometric distribution is the number of trials until the first success (or first failure) in Bernoulli trials. So, for example, asking for the probability that the number of tests until failure be at least n is the same as asking for the probability that the geometric is > n-1. A negative binomial is the sum of geometrics, so would be used, for example, when asking about the number of trials needed until detection of the 4th failure: that would be the sum of 4 geometrics. The geometric distribution with parameter p is $P\{X=k\} = p q^{k-1} \text{ for } k = 1, 2, \ldots,$ where q = 1-p. A sum of n geometric random variables has distribution
$$P\{X=k\}= (-1)^{k-n} {-n\choose k-n} p^n q^{k-n} = {k-1 \choose k-n} p^n q^{k-n}, \; k = n, n+1, \ldots.$$ This does involve the "negative binomial" C(-n,k-n).

RGV

Last edited: Feb 21, 2012
19. Feb 21, 2012

### Dick

Ok, so I just picked an overly general case? I wasn't actually sure what it was called and picked the first distribution I found that seemed to fit. Thanks!

20. Feb 21, 2012

### MarcoD

Looked all okay to me. I am stuck on proving why the variance is pr/(1-p)^2.

A. 0.9
B. 0.009
C. 1/9
D. 10/81

No, I guess D is the other one.

Last edited by a moderator: Feb 21, 2012