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# A certain candy store sells jellybeans in the following six

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A certain candy store sells jellybeans in the following six [#permalink]  01 Aug 2018, 10:30
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Question Stats:

28% (01:08) correct 72% (01:31) wrong based on 25 sessions
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. $$\frac{3}{4}$$
[Reveal] Spoiler: OA

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Re: A certain candy store sells jellybeans in the following six [#permalink]  03 Aug 2018, 21:10
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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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Re: A certain candy store sells jellybeans in the following six [#permalink]  18 Nov 2018, 18:03
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Re: A certain candy store sells jellybeans in the following six [#permalink]  18 Nov 2018, 20:02
Expert's post
AE wrote:

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. $$\frac{3}{4}$$

Let us choose that none are grape jellybeans..
1) when the box contains 2 flavours -
total ways - $$T_1=6C2=15$$ and without jelly beans - $$W_1=5C2=10$$
2) when the box contains 3 flavours -
total ways - $$T_2=6C3=20$$ and without jelly beans - $$W_2=5C3=10$$
3) when the box contains 4 flavours -
total ways - $$T_2=6C4=15$$ and without jelly beans - $$W_2=5C4=5$$

Probability that jelly is not there = $$\frac{W_1+W_2+W_3}{T_1+T_2+T_3}=\frac{10+10+5}{15+20+15}=\frac{25}{50}=\frac{1}{2}$$
so Probability that jelly is there = $$1-\frac{1}{2}=\frac{1}{2}$$

D
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Some useful Theory.
1. Arithmetic and Geometric progressions : https://greprepclub.com/forum/progressions-arithmetic-geometric-and-harmonic-11574.html#p27048
2. Effect of Arithmetic Operations on fraction : https://greprepclub.com/forum/effects-of-arithmetic-operations-on-fractions-11573.html?sid=d570445335a783891cd4d48a17db9825
3. Remainders : https://greprepclub.com/forum/remainders-what-you-should-know-11524.html
4. Number properties : https://greprepclub.com/forum/number-property-all-you-require-11518.html
5. Absolute Modulus and Inequalities : https://greprepclub.com/forum/absolute-modulus-a-better-understanding-11281.html

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Re: A certain candy store sells jellybeans in the following six [#permalink]  08 Jan 2019, 23:56
Expert's post
chetan2u wrote:
AE wrote:

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. $$\frac{3}{4}$$

Let us choose that none are grape jellybeans..
1) when the box contains 2 flavours -
total ways - $$T_1=6C2=15$$ and without jelly beans - $$W_1=5C2=10$$
2) when the box contains 3 flavours -
total ways - $$T_2=6C3=20$$ and without jelly beans - $$W_2=5C3=10$$
3) when the box contains 4 flavours -
total ways - $$T_2=6C4=15$$ and without jelly beans - $$W_2=5C4=5$$

Probability that jelly is not there = $$\frac{W_1+W_2+W_3}{T_1+T_2+T_3}=\frac{10+10+5}{15+20+15}=\frac{25}{50}=\frac{1}{2}$$
so Probability that jelly is there = $$1-\frac{1}{2}=\frac{1}{2}$$

D

This explanation was amazing! A real Eye opener!
Re: A certain candy store sells jellybeans in the following six   [#permalink] 08 Jan 2019, 23:56
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