Finite math

Question

One cage contains 4 white mice and 3 black ones. Another one contains 4 white mice and 2 black ones. A cage is chosen at random and three nice are selected. Find the expected number of white mice in the sample. 

Answer 1

This is similar to your other question but a bit harder.

 

cage X: 4W  3B      Cage Y: 4W   2B

 

First find the expected number of white mice if cage X is selected and do the same for cage Y

 

cage X:

     #W    P(of that #)

================ the denominator represent the number of ways of choosing 3 from 7

      0      [C(4,0)C(3,3)] / C(7,3)  the numerator  represents the # of ways of choosing 0W + 3B

      1      [C(4,1)C(3,2)] / C(7,3)  the numerator  represents the # of ways of choosing 1W + 2B

      2      [C(4,2)C(3,1)] / C(7,3)   2W + 1B

      3      [C(4,3)C(3,0)] / C(7,3)   3W    0B

 

#W is your random variable x so again use the formula Σx(px) to get expected value of black mice from cage X

 

Do the same for cage Y but change the numbers to reflect 4W and 2B (careful only 2B)

 

That gives you EXP(X) and EXP(Y). Since you are only picking one cage multiply each of these by 1/2 (1/2 because you're picking a cage at random) and then add the two results. So the final result would be calculated as: 

EXP(W) = (1/2)EXP(X)+(1/2)EXP(Y)

 

Need more help add a comment.

 

 

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