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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] New post 01 Aug 2018, 10:30
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30% (01:08) correct 69% (01:34) wrong based on 23 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] New post 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] New post 18 Nov 2018, 18:03
Please provide a short method.
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Re: A certain candy store sells jellybeans in the following six [#permalink] New post 18 Nov 2018, 20:02
Expert's post
AE wrote:
Please provide a short method.

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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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] New post 08 Jan 2019, 23:56
chetan2u wrote:
AE wrote:
Please provide a short method.

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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A certain candy store sells jellybeans in the following six

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