Carcass wrote:
A certain candy store sells jellybeans in the following six flavors only: banana, chocolate, grape, lemon, peach and strawberry. The jellybeans are sorted into boxes containing exactly 2, 3 or 4 different flavors, with each possible assortment of flavors appearing in exactly one box. What is the probability that any given box contains grape jellybeans?
A. \(\frac{1}{6}\)
B. \(\frac{1}{3}\)
C. \(\frac{2}{5}\)
D. \(\frac{1}{2}\)
E. \(\frac{3}{4}\)
Here,
In this type we find the probability of no grape then 1 - probability of no grape will give the answer
To start with let us take the first option of the box containing 2 flavors:Therefore the probability of box containing 2 flavors = \(\frac{1}{3}\)(Since there are 3 box containing 2 flavors, 3 flavors or 4 flavors)
Now probability of no grape in that box = \(\frac{5}{6} * \frac{4}{5} = \frac{2}{3}\).
Therefore the probability of having 2 flavors and no grape = \(\frac{1}{3} * \frac{2}{3}= \frac{2}{9}\).
the probability of box containing 3 flavors = \(\frac{1}{3}\)(Since there are 3 box containing 2 flavors, 3 flavors or 4 flavors)Probability of no grape in the 3 flavor box = \(\frac{5}{6} * \frac{4}{5} * \frac{3}{4} = \frac{1}{2}\).
Therefore the probability of having 2 flavors and no grape = \(\frac{1}{3} * \frac{1}{2}= \frac{1}{6}\).
the probability of box containing 4 flavors = \(\frac{1}{3}\)(Since there are 3 box containing 2 flavors, 3 flavors or 4 flavors)Probability of no grape in the 4 flavor box = \(\frac{5}{6} * \frac{4}{5} * \frac{3}{4} * \frac{2}{3} =\frac{1}{3}\).
Therefore the probability of having 4 flavors and no grape = \(\frac{1}{3} * \frac{1}{3} = \frac{1}{9}\).
Therefore probability of no grape in 2, 3 or 4 flavor box = \(\frac{2}{9} + \frac{1}{6} + \frac{1}{9} = \frac{1}{2}\).
Therefore the probability of having grape in any given box = \(1 - \frac{1}{2} = \frac{1}{2}\).
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