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.(adsbygoogle = window.adsbygoogle || []).push({});

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. [tex]A_i[/tex] would indicate that at least one [tex]i[/tex] was drawn, such that [tex]4-T = A_1 + A_2 + A_3+A_4[/tex]. Expectation is linear, so [tex] E(T) = 4 - E(A_1) -E(A_2) - E(A_3)-E(A_4)[/tex]. The expectation of $A_i$ is just the probability of obtaining it, so [tex]E(t) = 4 - 3(1/7+1/7+2/7+3/7) = 1[/tex]

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?

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Expectations of Random Variables

**Physics Forums | Science Articles, Homework Help, Discussion**