document.write( "Question 1081622: URN 1 contains 4 red balls and 3 black balls. URN 2 contains 5 red balls and 3 black balls. URN 3 contains 2 red and 3 black. If an urn is selected at random and a ball is drawn, find the probability that it will be red. \n" ); document.write( "
Algebra.Com's Answer #695685 by Boreal(15235)![]() ![]() You can put this solution on YOUR website! P(urn 1)=(1/3) and probability urn 1 and red ball is (4/7)(1/3)=(4/21) \n" ); document.write( "P(urn 2)=(1/3) and probability urn 2 and red ball is (5/8)(1/3)=(5/24) \n" ); document.write( "P(urn 3)=(1/3) and probability urn 3 and red ball is (2/5)(1/3)=(2/15) \n" ); document.write( "The sum of those probabilities has common denominator 21*15*24 \n" ); document.write( "numerator is 4*(15*24)+5(21*15)+2(21*24)/(21*15*24) \n" ); document.write( "numerator is 1440+1575+1008=4023 \n" ); document.write( "denominator is 7560 \n" ); document.write( "probability is 0.5321 ANSWER \n" ); document.write( "by adding 4/21(0.190) and 5/24 (0.208) and 2/15(0.133), one gets 0.531, but there is rounding error. \n" ); document.write( " \n" ); document.write( " |