Does this problem make sense? (Probability and Survey Sampling)

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SUMMARY

The discussion revolves around a probability question related to survey sampling, specifically focusing on the calculation of probabilities when sampling without replacement. The user expresses confusion regarding the probabilities of selecting elements on subsequent picks after the first selection, particularly in Plan 2 of the assignment. The correct approach involves using conditional probabilities, as demonstrated by the formula P(S[2]=i)=P(S[2]=i|S[1]≠i)*P(S[1]≠i). This method clarifies how to compute the probabilities for the second selection based on the outcome of the first.

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gjones89
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Hi,

I am trying to answer a question for an assignment, but am convinced that it doesn't make sense. Please see here: http://img813.imageshack.us/img813/1585/qst.png

My problem is with Plan 2 in the question. The probability of selecting any of the 5 elements on the first selection is 1/5. So for example, I can select element 1 with probability 1/5 on the first pick. But then, how can I work out the probabilities of choosing the 2nd, 3rd, 4th and 5th elements on the 2nd pick? Obviously these probabilities can't just be the numbers in the last column in the table, because if I add together the probabilties for choosing elements 2, 3, 4, and 5 (ignoring row 1 because we are sampling without replacement), I get (1/4 + 1/2 + 1/5 + 7/40) = 9/8, i.e. a total greater than one!

What is going on here? I want to work out the probability of picking, for example, element 1 on the first pick and element 2 on the second pick. Can anyone help?

Thanks.
 
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If I understand problem correctly then
P(S[2]=i)=P(S[2]=i|S[1][tex]\neq[/tex]i)*P(S[1][tex]\neq[/tex]i)

so for i=1 then

P(S[2]=1)=(1/8)*(4/5)


will be the same method for other elements
 

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