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Broker receives an order to purchase N units of stock A and M units of stock B - if individual orders are placed at random, what is the probability that all N units of A are purchased first (and therefore all M units of B are purchased last).

My Answer: The probability 1 unit of A is purchased before M units of B is 1/M+1 and hence the probability that N units of A are purchased before M units of B will be (1/M+1)^N. Is this correct?

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  • $\begingroup$ Welcome to Cross Validated! If this is a homework question (or you want it to be treated like one), please add the self-study tag so you get appropriate answers! $\endgroup$ – Matt Krause Oct 16 '15 at 15:34
  • $\begingroup$ @MattKrause No its not actually. Im working on interview Questions. $\endgroup$ – Jojo Oct 16 '15 at 16:01
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It sounds like having a non-bernoullian extraction here where the prob. to extract A at first try is N/(N+M). The prob. to extract A again at second try is (N-1)/(N+M-1)and again and again. Assuming the joint probability to extract AAA...A N-times in a row as the multiplication of the above it should yield to:

[N (N-1) (N-2) ... 1] /[(N+M)(N+M-1) ... M]

which if I am not wrong should be

[N! (M-1)!]/(N+M)!

Sorry for bad formatting.

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