# Homework Help: Expectations of Random Variables

1. Jun 9, 2010

### hbweb500

I am working on correcting an exam so that I may study for my probability final. Unfortunately, I don't have the correct answers, so I was hoping that someone here might be able to check my thought process.

1) Pick three numbers without replacement from the set {1,2,3,3,4,4,4}. Let T be the number of values that do not appear in your sample. Find the expectation of T.

I believe that to solve this one would use counting indicator variables. $$A_i$$ would indicate that at least one $$i$$ was drawn, such that $$4-T = A_1 + A_2 + A_3+A_4$$. Expectation is linear, so $$E(T) = 4 - E(A_1) -E(A_2) - E(A_3)-E(A_4)$$. The expectation of $A_i$ is just the probability of obtaining it, so $$E(t) = 4 - 3(1/7+1/7+2/7+3/7) = 1$$

There are a couple of things that worry me. First, if we scale up the set so that in includes 100 elements: {1,2,3,3,4,4,4...,4,4}, then there are still 4 distinct values. I still choose three numbers at random. By the calculation above, I get 1 again, though I would expect it now to be closer to 3, since I have almost no chance of drawing a 1, 2, or 3. I think the fault lies in my calculation of the probabilities. I want the probability that I get *at least one* 1, at least one 2, and so one. I could do this by taking the complement: 1 minus the probability that I get no ones, no twos, etc, but this would be a complicated binomial problem, wouldn't it?

2. Jun 10, 2010

### lanedance

I've read through this, but can you explain "Let T be the number of values that do not appear in your sample"?

say C is your choice, do you mean if:
C = {4,4,4} then T = 3
C = {4,4,3} then T = 2 and so on?

3. Jun 10, 2010

### hbweb500

Yes, that is what I mean.

In any case, I thought about it, and the determination of the probabilities isn't as hard as I thought it would be. For example, the probability of getting at least one 2 is the complement of getting no twos.

The probability of getting no twos is:

$$\frac{{1 \choose 0} {6 \choose 3}}{{7 \choose 3}}$$

Which is the standard calculation for sampling without replacement.

4. Jun 10, 2010

### lanedance

i'm still thinking on it, but i think one bust might be the Ai are not purely independent events...

so its worth examining it from a case point of view & see if we can generalise from there... then we can pick the easiest cases to solve remembering the probs must all sum to 1.

T=4, clearly P(T=4) = 0

T=3
this can only happen if we choose 444,

T=2
this can happen with the following outcomes: 133, 144, 233, 244, 344, 334

T=1
that can happen if we draw: 123, 134, 234

T= 0, P(T=0) = 0, again clearly

so looking at the T=3 case, the number of the ways to draw 3 objects from 7 without any order is:
$${}^7 C_{3} = \frac{7!}{3!4!}$$
there is only distinct case where we choose all 3 4s, so the probabilty is:
$$Pr(T=3) = \frac{1}{{}^7 C_{3}}$$

5. Jun 10, 2010

### lanedance

then if you can do the probabilities for T=1, you've cracked it...

6. Jun 10, 2010

### lanedance

another case based way is the X| method, X means we take a number, | means we skip to the next pile

as said, in total there are ${}^7 C_{3} = \frac{7!}{3!4!} =7.5 = 35$ distinct combinations we can end up with, when order is not important

cases are

T = 3
|||XXX = 1

T = 2
X||XX| = 1
X|||XX = 3C2
|X||XX|= 1
|X|||XX = 3C2
||XX|X = 3C1
||X|XX = (2C1)(3C2)

T = 1
X|X|X| = 2C1
X||X|X = (2C1)(3C1)
|X|X|X = (2C1)(3C1)