Probability problem involving hyper-geometric distributionn balls are chosen randomly and without replacement from an urn containing N white balls and M black balls. Give the probability mass function of the random variable X which counts the number of white balls chosen. Show that the expectation of X is N n ( M + N ) .Hint: Do not use the hypergeometric distribution. Instead, write X = X 1 + X 2 + . . . + X N where X i equals 1 if the ith white ball was chosen.

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Probability problem involving hyper-geometric distributionn balls are chosen randomly and without replacement from an urn containing N white balls and M black balls. Give the probability mass function of the random variable X which counts the number of white balls chosen. Show that the expectation of X is             N      n              (      M      +      N      )      .Hint: Do not use the hypergeometric distribution. Instead, write   X  =      X    1    +      X    2    +  .  .  .  +      X    N   where       X    i   equals 1 if the ith white ball was chosen.

Adelaide Barr · Accepted Answer

Step 1  P  (      X    i    =  0  )  =            N      +      M      −      1              N      +      M        ⋅            N      +      M      −      2              N      +      M      −      1        ⋅        .  .  .        ⋅            N      +      M      −      n              N      +      M      −      (      n      −      1      )       which, when simplified gives:  P  (      X    i    =  0  )  =            N      +      M      −      n              N      +      M      The way to think about this is the following.       X    i    =  0 denotes the fact of not picking the ith white ball. So, in the n draws, we&#039;re allowed to pick any other ball. Therefore, during the first draw, we can pick   N  +  M  −  1 balls (as we don&#039;t want to pick the ith white ball), and the probability to do so is   p  =            N      +      M      −      1              N      +      M      .When we do the second draw we are allowed to pick   N  +  M  −  2 balls (as we can&#039;t pick the ith white ball AND since there&#039;s only   N  −  M  −  1 balls in the urn after the first draw),and the probability to do so is   p  =            N      +      M      −      2              N      +      M      −      1      Applying the same logic to the n draws we come to the conclusion that   P  (      X    i    =  0  )  =            N      +      M      −      n              N      +      M      Step 2Now, we have   P  (      X    i    =  1  )  =  1  −  P  (      X    i    =  0  )    =      n          M      +      N      .Finally, to get the expected value of X, first we notice that   E  (      X    i    )  =  P  (      X    i    =  1  ) (see post for more details), then from the linearity of the expected value, we have that  E  (  X  )  =  E  (      X    1    )  +  E  (      X    2    )  +  .  .  .  +  E  (      X    N    )    =      ∑          i      =      1              N        E  (      X    i    )    =            N      ⋅      n              N      +      M       which is the correct solution.

n balls are chosen randomly and without replacement from an urn containing N white balls and M black balls. Give the probability mass function of the random variable X which counts the number of white balls chosen. Show that the expectation of X is (Nn)/(M+N)

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