Re: Pure probability question
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But why did you write out the expansion for e, as though the statement below it followed directly from that expansion (in your original answer in white)?
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# draws necessary = Z1 + Z2 + Z3 + ...
where
Zi = 0 if the ith draw was not necessary
Zi = 1 if the ith draw was necessary (which happens with probability 1/(i-1)! )
The expected number of draws necessary is the sum of the probabilities of requiring the first draw (1/0!), the second draw (1/1!), the third draw (1/2!), etc.
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