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Consider a population with 10 elements , N:{0,1,2,3,4,5,6,7,8,9}.

The probability of drawing

s1   0        is 0.1
s2   1        is 0.1
s3   2,3,4,5  is 0.4
s4   6,7,8,9  is 0.4

the goal is to draw 2 samples without replacement. So these are the possible draws

 Sample
 {s1,s2}
 {s1,s3}
 {s1,s4}
 {s2,s3}
 {s2,s4}
 {s3,s4}

The probability of drawing sample {s1,s3} is given as follows

p(s1,s3) = p(s1 in the 1st draw and s3 in the 2nd draw) + 
           p(s3 in the 1st draw and s1 in the 2nd draw)

p(s1,s3) = 0.1*(0.4/ 0.9)  + 0.4*(0.1/0.6) = 0.1111

I am unable to understand the logic behind this formula. 0.1*(0.4/ 0.9) + 0.4*(0.1/0.6)

Since these are mutually exclusive and independent events I am estimating the p(s1,s3) to be

p(s1,s3) = p(s1 in the 1st draw and s3 in the 2nd draw) + 
           p(s3 in the 1st draw and s1 in the 2nd draw)

p(s1,s3) = 0.1*(0.4)  + 0.4*(0.1) = 0.08

I need help understanding why p(s1,s3) is 0.1111 and not 0.08 ? Thanks in advance.

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1 Answer 1

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Since you're sampling without replacement, the events are not independent. You have to condition the second event on outcome of the first one.

For example, p(s1 in the 1st draw and s3 in the 2nd draw) is calculated as p(s1 in the 1st draw) $*$ p(s3 in the 2nd draw | s1 in the first draw). This is because the probability of getting s3 on the 2nd draw changes after you've drawn s1 on the 1st draw. After you've drawn one number, if you don't replace it, there are only 9 numbers left, so the probability of drawing each number increases.

If you were sampling with replacement, the events would be independent and p(s1, s3) would be 0.08 as you describe.

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  • $\begingroup$ I need to dwell on this a bit more. Thanks for the insight. $\endgroup$
    – bison2178
    Jan 19, 2020 at 15:47

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