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Jellybean Probability

by pkw209 » Fri Apr 30, 2010 12:14 pm
A certain company sells jellybeans in the following six flavors only: banana, chocolate, grape, lemon, peach and strawberry. If the jellybeans are sorted randomly into boxes containing exactly 2, 3 or 4 different flavors only, what is the probability that any given box contains grape jellybeans?

a) 1/6

b) 1/3

c) 2/5

d) 1/2

e) 3/4
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by Testluv » Fri Apr 30, 2010 10:47 pm
Let's consider the probability of not having grape, and then subtract from 1.

For the boxes with 2 flavors:

Prob = #des/#total

5C2/6C2 = 2/3
(the denominator is all the ways we can pull out any 2 flavors from the 6; the numerator is all the ways we can pull out 2 non-grape flavors from the 5 availabe)

For the boxes with 3 flavors: 5C3/6C3 = 1/2

For the boxes with 4 flavors: 5C4/6C4 = 1/3

Because the jellybeans are randomly assigned to each of the 3 kinds of boxes, the probability that you have any particular one kind of box is 1/3.

Thus, the probability of NOT having grape is: (1/3)*(2/3) + (1/3)*(1/2) + (1/3)*(1/3) = 1/2. Thus, the probability of selecting grape is also 1/2.

Choose D.

__________

The calculations are greatly sped up if you know that 5C4 = 5C1 = 5 (nC1 = n). Or that 5C3 = 5C2. Or that 6C4 = 6C2. Let's take 5C3 as an example. Every time we select a trio from 5, we are also "setting aside" a pair. Thus, 5C3 = 5C2. Similarly, 50C30 = 50C20, or 100C75 = 100C25, etc.

__________

Rather than using combinatorics, we could have also considered order. The probability of NOT having grape in the boxes with 2 flavors is (5/6)*(4/5) = 2/3, and so on. (Here, you can solve either by ignoring order--my first approach--or by considering order--my second approach. It doesn't matter, so long as you are consistent.)
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by pkw209 » Mon May 03, 2010 4:17 pm
Because the jellybeans are randomly assigned to each of the 3 kinds of boxes, the probability that you have any particular one kind of box is 1/3.
Ah! Thank you! This problem was driving me crazy. I got up to the 2/3, 1/2, 1/3 respectively for 2, 3, or 4 different flavors and then added them all up without multiplying each by 1/3 to get a number greater than 1, which obviously couldn't be right.

Now I can sleep tonight. :)
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by goyalsau » Tue Nov 30, 2010 3:15 am
Testluv wrote: Because the jellybeans are randomly assigned to each of the 3 kinds of boxes, the probability that you have any particular one kind of box is 1/3.

I did not realize that only one box will have the jellybeans that's why getting the probability more than 1, That is for sure wrong,

Thanks for the wonderful explanation.
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by nipunkathuria » Tue Nov 30, 2010 4:53 pm
Testluv wrote:Let's consider the probability of not having grape, and then subtract from 1.

For the boxes with 2 flavors:

Prob = #des/#total

5C2/6C2 = 2/3
(the denominator is all the ways we can pull out any 2 flavors from the 6; the numerator is all the ways we can pull out 2 non-grape flavors from the 5 availabe)

For the boxes with 3 flavors: 5C3/6C3 = 1/2

For the boxes with 4 flavors: 5C4/6C4 = 1/3

Because the jellybeans are randomly assigned to each of the 3 kinds of boxes, the probability that you have any particular one kind of box is 1/3.

Thus, the probability of NOT having grape is: (1/3)*(2/3) + (1/3)*(1/2) + (1/3)*(1/3) = 1/2. Thus, the probability of selecting grape is also 1/2.

Choose D.

__________

The calculations are greatly sped up if you know that 5C4 = 5C1 = 5 (nC1 = n). Or that 5C3 = 5C2. Or that 6C4 = 6C2. Let's take 5C3 as an example. Every time we select a trio from 5, we are also "setting aside" a pair. Thus, 5C3 = 5C2. Similarly, 50C30 = 50C20, or 100C75 = 100C25, etc.

__________

Rather than using combinatorics, we could have also considered order. The probability of NOT having grape in the boxes with 2 flavors is (5/6)*(4/5) = 2/3, and so on. (Here, you can solve either by ignoring order--my first approach--or by considering order--my second approach. It doesn't matter, so long as you are consistent.)


Hi.. cant we approach this question in the following fashion:

The prob of it being grape=1/6
we have three boxes with 2,3, and 4 beans ...
hence :
Box 1: P(Grape)=1/6 P(other)=1(5/5)
Box 2: same
Box 3: Same
Hence P=P1+P2+P3
=1/6*1+1/6*1+1/6*1

=3/6

=1/2

This was what i finished with...any comments?
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by sumit.sinha » Wed Dec 08, 2010 12:35 pm
nipunkathuria wrote:
Testluv wrote:Let's consider the probability of not having grape, and then subtract from 1.

For the boxes with 2 flavors:

Prob = #des/#total

5C2/6C2 = 2/3
(the denominator is all the ways we can pull out any 2 flavors from the 6; the numerator is all the ways we can pull out 2 non-grape flavors from the 5 availabe)

For the boxes with 3 flavors: 5C3/6C3 = 1/2

For the boxes with 4 flavors: 5C4/6C4 = 1/3

Because the jellybeans are randomly assigned to each of the 3 kinds of boxes, the probability that you have any particular one kind of box is 1/3.

Thus, the probability of NOT having grape is: (1/3)*(2/3) + (1/3)*(1/2) + (1/3)*(1/3) = 1/2. Thus, the probability of selecting grape is also 1/2.

Choose D.

__________

The calculations are greatly sped up if you know that 5C4 = 5C1 = 5 (nC1 = n). Or that 5C3 = 5C2. Or that 6C4 = 6C2. Let's take 5C3 as an example. Every time we select a trio from 5, we are also "setting aside" a pair. Thus, 5C3 = 5C2. Similarly, 50C30 = 50C20, or 100C75 = 100C25, etc.

__________

Rather than using combinatorics, we could have also considered order. The probability of NOT having grape in the boxes with 2 flavors is (5/6)*(4/5) = 2/3, and so on. (Here, you can solve either by ignoring order--my first approach--or by considering order--my second approach. It doesn't matter, so long as you are consistent.)


Hi.. cant we approach this question in the following fashion:

The prob of it being grape=1/6
we have three boxes with 2,3, and 4 beans ...
hence :
Box 1: P(Grape)=1/6 P(other)=1(5/5)
Box 2: same
Box 3: Same
Hence P=P1+P2+P3
=1/6*1+1/6*1+1/6*1

=3/6

=1/2

This was what i finished with...any comments?
Yes, this is the method which even i used. Here we are not considering the probability of selecting one of the box. But this method also gives me the correct answer. Any explanations how come?? Is it correct or wrong approach?
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by junegmat221 » Tue Dec 14, 2010 7:19 am
Hi.. cant we approach this question in the following fashion:

The prob of it being grape=1/6
we have three boxes with 2,3, and 4 beans ...
hence :
Box 1: P(Grape)=1/6 P(other)=1(5/5)
Box 2: same
Box 3: Same
Hence P=P1+P2+P3
=1/6*1+1/6*1+1/6*1

=3/6
Experts please comment whether this can also be used.
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by tomada » Tue Jan 18, 2011 5:23 pm
I think that this just happens to be concidence that the answer is the same.
Suppose the question was rephrased so that only 2 jellybeans could exist in any particular box - that is, the choices of having 3 or 4 jellybeans in a box no longer exist.

Using the method below, the probability of having a grape jellybean in a box containing 2 jellybeans = 1/6*5/5 = 1/6.
However, we know that the probability of having a grape jellybean in a box containing 2 jellybeans = 1/3.

junegmat221 wrote:
Hi.. cant we approach this question in the following fashion:

The prob of it being grape=1/6
we have three boxes with 2,3, and 4 beans ...
hence :
Box 1: P(Grape)=1/6 P(other)=1(5/5)
Box 2: same
Box 3: Same
Hence P=P1+P2+P3
=1/6*1+1/6*1+1/6*1

=3/6
Experts please comment whether this can also be used.
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by anshumishra » Tue Jan 18, 2011 5:56 pm
tomada wrote:I think that this just happens to be concidence that the answer is the same.
Suppose the question was rephrased so that only 2 jellybeans could exist in any particular box - that is, the choices of having 3 or 4 jellybeans in a box no longer exist.

Using the method below, the probability of having a grape jellybean in a box containing 2 jellybeans = 1/6*5/5 = 1/6.
However, we know that the probability of having a grape jellybean in a box containing 2 jellybeans = 1/3.

junegmat221 wrote:
Hi.. cant we approach this question in the following fashion:

The prob of it being grape=1/6
we have three boxes with 2,3, and 4 beans ...
hence :
Box 1: P(Grape)=1/6 P(other)=1(5/5)
Box 2: same
Box 3: Same
Hence P=P1+P2+P3
=1/6*1+1/6*1+1/6*1

=3/6
Experts please comment whether this can also be used.
Right, that was coincidence that the answer matched.

This is how the probability method can be used :

Probability = (probability of selecting the 1st box AND probability of being grape jelly in that box) + (probability of selecting the 2nd box AND probability of being grape jelly in that box) + (probability of selecting the 3rd box AND probability of being grape jelly in that box)

= (1/3 * 2/6) + (1/3 * 3/6) + (1/3 * 4/6) = 1/3 * 9/6 = 9/18 = 1/2.

[Note : p(grape jelly in box 1) = 2/6 , as we have two jellys out of 6 , so probability of being any jelly is 2/6....similarly for the other cases..]
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by gmat1978 » Mon Mar 14, 2011 9:15 pm
Can anyone please help... I got lost midway.
Testluv wrote:Let's consider the probability of not having grape, and then subtract from 1.

For the boxes with 2 flavors:

Prob = #des/#total

5C2/6C2 = 2/3
(the denominator is all the ways we can pull out any 2 flavors from the 6; the numerator is all the ways we can pull out 2 non-grape flavors from the 5 availabe)

For the boxes with 3 flavors: 5C3/6C3 = 1/2

For the boxes with 4 flavors: 5C4/6C4 = 1/3
I followed up to this point, however, I am not able to understand the below part in which each of the probabilities (2/3, 1/2, 1/3) is multiplied by 1/3. I guess my question is why do we have to multiply by1/3?
Testluv wrote:
Because the jellybeans are randomly assigned to each of the 3 kinds of boxes, the probability that you have any particular one kind of box is 1/3.

Thus, the probability of NOT having grape is: (1/3)*(2/3) + (1/3)*(1/2) + (1/3)*(1/3) = 1/2. Thus, the probability of selecting grape is also 1/2.

Choose D.

__________
Thanks.
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by force5 » Tue Mar 15, 2011 12:15 am
yes did the same way too.... 1/6*1+1/6*1+1/6*1 = 1/2
i think its a genuine approach too.
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by kprabhala.mba » Tue Mar 15, 2011 6:38 pm
Consider each box individually:

2 jellybean box:
Number of ways of selecting 2 jellybeans out of 6 = 6C2
Number of ways of finding a grape jellybean out of the 2 selected jellybeans = 2C1
Therefore, probability of finding a grape jellybean in the 2 jellybean box: 2C1/6C2

Similarly, the probability for other two boxes can be calculated as 3C1/6C3 and 4C1/6C4.

The probability of selecting a 2 jellybean box or a 3 jellybean box or a 4 jellybean box with
a grape jellybean in it is, therefore,

2C1/6C2 + 3C1/6C3 + 4C1/6C4 = 1/2

Answer D.
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by GMATGuruNY » Wed Mar 23, 2011 12:03 am
Can anyone please help... I got lost midway.
I received a PM asking me to comment. My approach wouldn't be much different from Testluv's, but perhaps the following will help those who are still confused.

Let's determine P(no grape):

P(2 flavors) = 1/3 (Since the box could contain 2 flavors, 3 flavors, or 4 flavors, giving us a 1/3 chance of choosing a box with 2 flavors.)
P(no grape) = 5/6 * 4/5 = 2/3.
In order to choose a box with 2 flavors and no grape, both of the events above must happen. When we want events to happen together, we multiply the probabilities:
P(2 flavors and no grape) = 1/3 * 2/3 = 2/9.

P(3 flavors) = 1/3 (Since the box could contain 2 flavors, 3 flavors, or 4 flavors, giving us a 1/3 chance of choosing a box with 3 flavors.)
P(no grape) = 5/6 * 4/5 * 3/4 = 1/2.
In order to choose a box with 3 flavors and no grape, both of the events above must happen. When we want events to happen together, we multiply the probabilities:
P(3 flavors and no grape) = 1/3 * 1/2 = 1/6.

P(4 flavors) = 1/3 (Since the box could contain 2 flavors, 3 flavors, or 4 flavors, giving us a 1/3 chance of choosing a box with 4 flavors.)
P(no grape) = 5/6 * 4/5 * 3/4 * 2/3 = 1/3.
In order to choose a box with 4 flavors and no grape, both of the events above must happen. When we want events to happen together, we multiply the probabilities:
P(4 flavors and no grape) = 1/3 * 1/3 = 1/9.

Since any of the outcomes above will result in a box with no grape, we add the fractions:
P(no grape) = 2/9 + 1/6 + 1/9 = 1/2.

Since P(no grape) = 1/2, P(grape) = 1/2.

The correct answer is D.
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by GMATGuruNY » Wed Mar 23, 2011 10:33 am
pkw209 wrote:A certain company sells jellybeans in the following six flavors only: banana, chocolate, grape, lemon, peach and strawberry. If the jellybeans are sorted randomly into boxes containing exactly 2, 3 or 4 different flavors only, what is the probability that any given box contains grape jellybeans?

a) 1/6

b) 1/3

c) 2/5

d) 1/2

e) 3/4
Here's another approach that some might find more straightforward:

Total number of different 2-flavor packages = 6C2 = 15.
Total number of different 3-flavor packages = 6C3 = 20.
Total number of different 4-flavor packages = 6C4 = 15.
Total number of different packages = 15+20+15 = 50.

Total number of different 2-flavor packages with grape = 5. (Since there are 5 other flavors that can be combined with grape.)
Total number of different 3-flavor packages with grape = 10. (Since, of the remaining 5 flavors, there are 5C2 = 10 pairs that can be combined with grape.)
Total number of different 4-flavor packages with grape = 10. (Since, of the remaining 5 flavors, there are 5C3 = 10 combinations of 3 that can be combined with grape.)
Total number of different packages with grape = 5+10+10 = 25.

(Total number of different packages with grape)/(Total number of different packages) = 25/50 = 1/2.

The correct answer is D.
Last edited by GMATGuruNY on Fri Jan 15, 2016 4:22 am, edited 1 time in total.
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by ngkontos » Wed Mar 23, 2011 4:49 pm
Hi guys,

Would it be wrong to solve as follows or was it just coincidence that I got the same answer?

Probability of getting a grape = 1/6
Number of different ways you can get a grape in:
2 pack: 2C1 = 2
3 pack: 3C1 = 3
4 pack: 4C1 = 4

Probability of getting grape in any of the 3 packs = (1/6)*(2) + (1/6)*3 + (1/6)*4 = 3/2. So obviously it cant be more than 100% but I thought that what i forgot to do is multiply by the probability of picking any of the 3 boxes which is 1/3 so that:

3/2 * 1/3 = 1/2, which is the correct answer

Is that the correct methodology or just a fluke? If it's correct then can someone please explain why you have to actually take into account the probability of picking just one of the 3 packs (ie last step)? i thought since any of 3 can include the grape you just have to add the individual probabilities.

Many thanks!
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