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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