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Feb
2
comment Expected value with indicator random variables
In case anyone was still wondering why P(object i is chosen by A) is 3/10: $$ $$ If A chooses 3 out of 10, the sample space is the set of all 3-sets chosen from 10: $$ {10 \choose 3} $$ And the number of these sets that will contain the 'i' of interest is [first pick 'i' and then out of the remaining 9 you can choose any two] : $$ 1 { 9 \choose 2 } $$ So the final probability is: $$ \frac{9 \choose 2}{10 \choose 3} = 3/10 $$
Jan
31
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