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)\"\" \"About 
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( "\"%281%2F60%29%2F%28%281%2F60%29%2B%2824%2F60%29%29\"
\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( "\"1%2F%281+%2B+24%29\"
\n" ); document.write( "
\n" ); document.write( "So...the final answer is 1/25.
\n" ); document.write( "
\n" );