SOLUTION: One cage contains 3 white mice and 2 black ones, and another cage contains 5 white mice and 2 black ones. A cage is chosen at random and three mice are selected. Find the expecte
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-> SOLUTION: One cage contains 3 white mice and 2 black ones, and another cage contains 5 white mice and 2 black ones. A cage is chosen at random and three mice are selected. Find the expecte
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Question 1026011: One cage contains 3 white mice and 2 black ones, and another cage contains 5 white mice and 2 black ones. A cage is chosen at random and three mice are selected. Find the expected number of white mice in the sample. Answer by robertb(5830) (Show Source):
You can put this solution on YOUR website! Assume that each cage is equally likely to be chosen.
Let X = the number of white mice in the three mice chosen
Apparently, x = 0, 1, 2, 3.
Given the first cage is chosen, the following are the conditional probabilities for each value of X.
X -----> 0 1 2 3
p(x|1st cage)---> 0 3/10 6/10 1/10
Given the second cage is chosen, the following are the conditional probabilities for each value of X.
X -----> 0 1 2 3
p(x|2nd cage)---> 0 1/7 4/7 2/7
This means the combined (joint) probability mass function is as follows:
X -----> 0 1 2 3
p(x)---> 0 62/280 164/280 54/280
==> E(X) =
Thus, the expected number of white mice in the selection is practically 2.
