Binomial Probability Clarification

[exitwound]

Homework Statement

For fixed n, are there values of p (0≤p≤1) for which V(X) = 0? Explain why this is so.

The Attempt at a Solution

For X~Bin(n,p), V(X)=np(1-p). So the only solutions for this equation are when p=0 or p=1. There are too many variables for me to keep track of and understand. Is p the probability of X occurring? And n is the number of trials in the experiment? If so, if the probability of X occurring is zero or one, how does that relate to variance of the outcomes of the experiments? I'm confused.

 

[Dick]

Can you explain roughly what variance means in words? And why p=0 and p=1 giving you zero variance makes perfect sense? Look up a nonmathematical definition of 'variance'.

 

[exitwound]

Variance is the amount of...dispersion of results around a certain point. I don't know how p and V(X) are related other than the equation. I know p is the probability of {success} of ... something. I don't know what that something is though. Is it X? I'm totally confused on what all the symbols in Binomial Probability mean.

 

[Dick]

Ok, yes, p is the probability of "success". If you do n trials and p=0 how many times do you succeed? Same for p=1. Is there any 'dispersion of results'?

 

[exitwound]

Say I roll a die and the probability of getting a 1 is 0%, (due to the die being weighted) which would mean that the probabiltity of not getting a 1 is 100%. If I do n, all of my results are 1, therefore the variance is 0. And if the probability of getting a 1 is 100%, then all results are 1, and again no variance. I didn't realize this until I wrote out this experiment actually, coincidentally forgetting that we're dealing with dichotomous experiments. Is it that easy?

 

[Dick]

Saying "dichotomous experiments" makes it sound hard. But, yes, it's that easy. If p=0 or p=1 then you know exactly what will happen. There is no dispersion. If p=(1/2) or anything else not 0 or 1 then you can get anywhere from 0 to n successes. There is dispersion and nonzero variance, just as the formula np(p-1) tells you.

 

[exitwound]

The second half of the problem asks: When is V(X) maximized? If we take the derivative and set it equal to zero, we get 1-2p=0 which means p=.5 which says that the maximum variance occurs when p=.5. This is where I get confused.

If the expected value of the Binomial is E(X)=np, and the variance is V(X)=np(1-p), I am not quite sure I picture what's happening. When p=.5, what is actually going on?

 

[Dick]

You are doing the math part right. If p is high, you 'usually' succeed, if p is low you 'usually' don't. With p=0 and p=1 being extreme cases of that. Doesn't it make sense that p=(1/2) is the maximum uncertainty? Hence maximum variance? These are all just words. The fact is that V(1/2)=n/4 is the largest value of the variance you get for any value of p. Try p=1/4 or p=3/4. Don't you trust you own math?

 

[exitwound]

I can "do" the math but I don't really understand it. When p=1/4, V(X)=n(1/4)(3/4)... when p=3/4, V(X)=n(3/4)(1/4)...which is the same. But I'm not really able to 'picture' what the variance is doing. For instance, if my possibilities of outcomes is 0 or 1, and the probability of getting the 1 is (1/4), why is the variance around the expected value any different than if it's 1/2, or 3/5, if the only two outcomes are 0 and 1?

 

[Dick]

Because 3n/8 is less than n/4. That's why. Graph V as a function of p. It peaks at p=1/2 doesn't it? Look up web apps that will show you graphs of binomial distributions for various values of n and p if you can't 'picture' it. For fixed n the value of p that gives you the biggest spread in outcomes is p=1/2.