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Given an urn with a number of two objects, $A$'s and $B$'s, if I am to find the probability of the $i$-th object drawn without replacement to be an $A$ would I need to compute all the different ways that $i-1$ objects can first be drawn?

For example if $i = 3$ would I need to first compute the probability that the first two objects drawn are $A$ then $A$, $A$ then $B$, $B$ then $A$, and $B$ then $B$. Then find the probability of the 3rd object drawn being $A$ in each of these instances and sum all 4 ways that the 3rd objects is $A$? My particular $i$ is 6 so I want to make sure this is correct before actually doing it.

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  • $\begingroup$ This is the right approach. $\endgroup$ Jan 27, 2014 at 4:36
  • $\begingroup$ @ChrisK Was afraid of that lol. Thanks for the confirmation. $\endgroup$
    – Alex
    Jan 27, 2014 at 4:44

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Suppose there are $a$ A's and $b$ B's. Then the probability that the $i$-th object drawn is an A is $\frac{a}{a+b}$.

You do not need to consider the various orders that A's and B's could come in on the first $i-1$ trials. One way to see this is to put ID numbers on all the A's and B's, to make them distinct.

Then any object is just as likely to be the $i$-th one drawn as any other. For all permutations of the objects are equally likely. So the probability that any particular A is drawn in the $i$-th position is $\frac{1}{a+b}$, and therefore the probability an A is in the $i$-th position is $\frac{a}{a+b}$.

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  • $\begingroup$ Isn't this for drawing $i$ objects with replacement? $\endgroup$
    – Alex
    Jan 27, 2014 at 5:08
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    $\begingroup$ It is for drawing without replacement. For fun, you might take say $a=4$ and $b=7$. Find the probability an A is second the long way, and simplify. You will get $\frac{4}{11}$. Find the probability an A is third the long way, and simplify. You will get $\frac{4}{11}$. $\endgroup$ Jan 27, 2014 at 5:12

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