SOLUTION: A bag contains 100 pieces of candy. More than 50 are cherry, the rest are grape. 2 pieces are picked simultaneously. The probability of drawing two pieces of the same flavor is the

Algebra ->  Statistics  -> Binomial-probability -> SOLUTION: A bag contains 100 pieces of candy. More than 50 are cherry, the rest are grape. 2 pieces are picked simultaneously. The probability of drawing two pieces of the same flavor is the      Log On


   

Question 1021467: A bag contains 100 pieces of candy. More than 50 are cherry, the rest are grape. 2 pieces are picked simultaneously. The probability of drawing two pieces of the same flavor is the same as the probability of drawing 2 pieces of different flavors. How many cherry flavored pieces are there?
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Let's say there are X cherry candies.
Then there are 100-X grape candies.
The probability of choosing CC or GG is,
P=P%28CC%29%2BP%28GG%29
P=%28%28X%29%2F100%29%28%28X-1%29%2F99%29%2B%28%28100-X%29%2F100%29%28%2899-X%29%2F99%29%29
P=%28X%28X-1%29%2B%28100-X%29%2899-X%29%29%2F9900
The probability of choosing CG or GC is,
P=P%28CG%29%2BP%28GC%29
P=%28X%2F100%29%28%28100-X%29%2F99%29%2B%28100-X%29%2F100%29%28X%2F99%29
P=%282X%28100-X%29%29%2F9900
So then make them equal to each other,
X%28X-1%29%2B%28100-X%29%2899-X%29=2X%28100-X%29
X%5E2-X%2B9900-199X%2BX%5E2=200X-2X%5E2
4X%5E2-400X%2B9900=0
X%5E2-100X%2B2475=0
%28X-55%29%28X-45%29=0
Two solutions but the question stipulated X%3E50
So then,
X=55 <--- Cherry
100-X=45 <--- Grape