SOLUTION: can someone please help me solve this? thanks At a certain pizza parlor,36 % of the customers order a pizza containing onions,35 % of the customers order a pizza containing saus

Question 185521: can someone please help me solve this? thanks
At a certain pizza parlor,36 % of the customers order a pizza containing onions,35 % of the customers order a pizza containing sausage, and 66% order a pizza containing onions or sausage (or both). Find the probability that a customer chosen at random will order a pizza containing both onions and sausage.

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
At a certain pizza parlor,36 % of the customers order a pizza containing onions,35 % of the customers order a pizza containing sausage, and 66% order a pizza containing onions or sausage (or both). Find the probability that a customer chosen at random will order a pizza containing both onions and sausage.
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Draw two intersecting circles.
Label one of the circles onions and 36%
Label the other as sausage and 35%
Label the intersection of the circles as "x".
------------------------------------------
Equation:
(36-x) is the percent that order onions only
(35-x) is the percent that order sausage only
x is the percent that order both
--------------------------------------
Equation:
(36-x) + (35-x) + x = 60
71 - x = 60
x = 11
So, the probability a randomly selected person has ordered onion and
sausage is 0.11
======================
Cheers,
Stan H.
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