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A certain candy store sells jellybeans in the following six
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01 Aug 2018, 10:30
01 Aug 2018, 10:30
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Question Stats:
42% (02:18) correct
57% (02:12) wrong based on 123 sessions
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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. \(\frac{3}{4}\)
A. \(\frac{1}{6}\)
B. \(\frac{1}{3}\)
C. \(\frac{2}{5}\)
D. \(\frac{1}{2}\)
E. \(\frac{3}{4}\)
Re: A certain candy store sells jellybeans in the following six
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18 Nov 2018, 20:02
18 Nov 2018, 20:02
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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. \(\frac{3}{4}\)
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
Re: A certain candy store sells jellybeans in the following six
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03 Aug 2018, 21:10
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. \(\frac{3}{4}\)
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 = \(1 - \frac{1}{2} = \frac{1}{2}\).
