document.write( "Question 1101386: Two balls are drawn in succession without replacement from a box containing 4 red balls and 6 blue balls. Let Z be the random variable representing the number of blue balls. Construct the probability distribution of the random variable Z. \n" ); document.write( "
Algebra.Com's Answer #715967 by Edwin McCravy(20060)![]() ![]() You can put this solution on YOUR website! \r\n" ); document.write( "First determine the probabilities of the events.\r\n" ); document.write( "\r\n" ); document.write( " events Probability\r\n" ); document.write( " RR (4/10)(3/9) = 2/15\r\n" ); document.write( " RB (4/10)(6/9) = 4/15\r\n" ); document.write( " BR (6/10)(4/9) = 4/15 \r\n" ); document.write( " BB (6/10)(5/9) = 1/3\r\n" ); document.write( "\r\n" ); document.write( "The probability of 0 blue balls (RR)is 2/15\r\n" ); document.write( "The probability of 1 blue ball is (RB or BR) is 4/15+4/15 = 8/15\r\n" ); document.write( "The probability of 2 blue balls (BB) is 1/3\r\n" ); document.write( "\r\n" ); document.write( "So the probability distribution is:\r\n" ); document.write( "\r\n" ); document.write( "Z p(Z)\r\n" ); document.write( "-------- \r\n" ); document.write( "0 2/15\r\n" ); document.write( "1 8/15\r\n" ); document.write( "2 1/3\r\n" ); document.write( "\r\n" ); document.write( "Notice that the sum of the probabilities = 2/15+8/15+1/3 = 1\r\n" ); document.write( "\r\n" ); document.write( "Edwin\n" ); document.write( " |