document.write( "Question 1141624: A bowl contains a total of 60 Bartlett, Bose and Anjou pears. The probability of randomly picking out a Barrtlett pear is 2/5 and the probability of picking out a Bose pear is 7/12. If all the Bose pears are removed, what is the probability of randomly picking An Anjou pear? \n" ); document.write( "
Algebra.Com's Answer #762201 by VFBundy(438)![]() ![]() You can put this solution on YOUR website! Probability of picking an Anjou pear (before Bose pears are removed): \n" ); document.write( " \n" ); document.write( "1 - 2/5 - 7/12 \n" ); document.write( " \n" ); document.write( "Simplify, using common denominator of 60: \n" ); document.write( " \n" ); document.write( "60/60 - 24/60 - 35/60 = 1/60 \n" ); document.write( " \n" ); document.write( "So...the odds of picking each pear are as follows: \n" ); document.write( " \n" ); document.write( "Bartlett = 24/60 \n" ); document.write( "Bose = 35/60 \n" ); document.write( "Anjou = 1/60 \n" ); document.write( " \n" ); document.write( "If you remove the Bose pears, then the odds of picking an Anjou pear are: \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "You can get rid of the denominator of 60 in each number so you are left with the much more manageable: \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "So...the final answer is 1/25. |