SOLUTION: 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

Algebra ->  Statistics  -> Normal-probability -> SOLUTION: 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      Log On


   

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?
Answer by VFBundy(438) About Me  (Show Source):
You can put this solution on YOUR website!
Probability of picking an Anjou pear (before Bose pears are removed):

1 - 2/5 - 7/12

Simplify, using common denominator of 60:

60/60 - 24/60 - 35/60 = 1/60

So...the odds of picking each pear are as follows:

Bartlett = 24/60
Bose = 35/60
Anjou = 1/60

If you remove the Bose pears, then the odds of picking an Anjou pear are:

%281%2F60%29%2F%28%281%2F60%29%2B%2824%2F60%29%29

You can get rid of the denominator of 60 in each number so you are left with the much more manageable:

1%2F%281+%2B+24%29

So...the final answer is 1/25.