- #1
hbweb500
- 41
- 1
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. [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?
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?