document.write( "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. \n" ); document.write( "
Algebra.Com's Answer #641271 by robertb(5830)![]() ![]() You can put this solution on YOUR website! Assume that each cage is equally likely to be chosen. \n" ); document.write( "Let X = the number of white mice in the three mice chosen \n" ); document.write( "Apparently, x = 0, 1, 2, 3. \n" ); document.write( "Given the first cage is chosen, the following are the conditional probabilities for each value of X. \n" ); document.write( "X -----> 0 1 2 3 \n" ); document.write( "p(x|1st cage)---> 0 3/10 6/10 1/10\r \n" ); document.write( "\n" ); document.write( "Given the second cage is chosen, the following are the conditional probabilities for each value of X. \n" ); document.write( "X -----> 0 1 2 3 \n" ); document.write( "p(x|2nd cage)---> 0 1/7 4/7 2/7\r \n" ); document.write( "\n" ); document.write( "This means the combined (joint) probability mass function is as follows:\r \n" ); document.write( "\n" ); document.write( "X -----> 0 1 2 3 \n" ); document.write( "p(x)---> 0 62/280 164/280 54/280\r \n" ); document.write( "\n" ); document.write( "==> E(X) = \n" ); document.write( "\n" ); document.write( "Thus, the expected number of white mice in the selection is practically 2. \n" ); document.write( " |