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?

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.