Simply, 2 options out of 6 is 2/6 = 1/3
Simply, 3 options out of 6 is 3/6 = 1/2
Simply, 4 options out of 6 is 4/6 = 2/3

Since every box has the same probability to be chosen

it's an average (1/3 + 1/2 + 2/3) / 3 = 1/2

I did this in the below way:

For the first box, the probability of choosing the 2 flavors out of 6 is 2/6 and for the grape to be one of them it should be 1/2.- so grape to be in the first box- (2/6)(1/2)= 1/6
Similarly for the other boxes the probably for the grape to be present is 1/6.

Total probability = (1/6)+(1/6)+(1/6) = 1/2.

Is this a right way?

total flavour = 6
3 boxes A,B,& C

A contains 2 flavours
2 can be selected out of 6 in 6C2 wauys = 15 ways
1 can be out of 5 and the other can be Grape = 5 ways

Probability :- 5 / 15 = 1/3 ways

B contains 3 falvours
Total - 6C3= 20 ways
1 is Grape, and 2 out of 5 can be selected in 5C2= 10
Probability = 12/20 = 1/2 ways


C contains 4 flavours

total 6C4 = 15 ways
1 is grape, other 3 can be selected out of 5 in 5C3 = 10 ways
Probability: - 10/15 = 2/3 ways

Also, we can have only one type of box at a time s0 -- any one type of box can be selected out of 3 in = 1/3 ways

ANSWER = 1/3 [(2/3)+(1/3)+(1/2)] = 1/2

Option D

Ans is 1/2 as explained by Mitch earlier. I followed the same approach. Good question.

Good explanation.

good question. One important point to remember is that out of three boxes we are inly selecting on in the end. I got 45/30 and was wondering whatto do,
But then we have to multiply by 1/3 and hence P = 1/2 answer comes.

Option D is correct

[quote="Testluv"]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.

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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.)[/quote]
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Using combinatorial here suggests that a box cannot have multiple jellybeans of a single flavor. If a box with 2 jellybeans has both jellybeans of grape flavor that would meet the criteria too...correct?

So shouldn't the total number of ways we can have 2 jellybeans will actually be 6*6 and not 6C2?

Am I missing something here? I am getting 1/6 as my answer (by allowing repetitions) and I am not entirely convinced that using combinations is the right approach since this problem does involve repetitions. Please explain your technique further.

Using combinatorial here suggests that a box cannot have multiple jellybeans of a single flavor. If a box with 2 jellybeans has both jellybeans of grape flavor that would meet the criteria too...correct?

So shouldn't the total number of ways we can have 2 jellybeans will actually be 6*6 and not 6C2?

Am I missing something here? I am getting 1/6 as my answer (by allowing repetitions) and I am not entirely convinced that using combinations is the right approach since this problem does involve repetitions. Please explain your technique further.

there are a total of 6 flavors
1) we need 2 out of 6 = 2/6
and probability of a grape jelly bean present in those 2 flavors = 1/2
2/6 * 1/2 = 1/6

2) we need 3 flavors out of 6 = 3/6
and one grape to be present in those 3 selected flavors = 1/3
3/6 * 1/3 = 1/6

3) we need 4 flavors = 4/6
and one grape to be present in those 4 flavors = 1/4
4/6*1/4 = 1/6

the grape can be present in (1) or (2) or (3)
so, 1/6 + 1/6 + 1/6 = 3/6 = 1/2

Ans D

Note: remember AND means MULTIPLICATION "*"
OR means ADDITION "+"

D is the answer.

I am confused,
I am trying to know the probability of a grape jelly bean present in each box.
Can anyone help me? I dont know wich way is de correct one:


1) Probability of a grape gb present in BOX1: 2/6

2) Probability of a grape gb present in BOX1 : 1/6 + 1/5

Thanks!

As the question does not say how many of each flavour there are, we cannot know for certain what the probabilities are.

E.g. There could be 1 grape jellybean and 1000 of each of the others, or 1000 grape jellybeans and 1 of each of the others. INSUFFICIENT.

GMATguruNY:
The solution which you have given earlier mentions selection of the boxes, but the question never asks about selecting any box. it only asks for the prob. that any box contains grape. so we need not to go thru this lengthy process.

we can simply calculate the probabilities of having grape JB in all of the boxes individually and then adding it.
so we get 1/6 in each case.
total p(2,3,4)= P(2) +P(3) +P(4)= 3*(1/6) =1/2 =D

romitrock wrote:we can simply calculate the probabilities of having grape JB in all of the boxes
we get 1/6 in each case.

The value in red is incorrect.
To illustrate, here are all of the possible 2-color boxes that can be formed from the 6 colors banana, chocolate, grape, lemon, peach and strawberry:
BC, BG, BL, BP, BS, CG, CL, CP, CS, GL, GP, GS, LP, LS, PS.
Of the 15 possible 2-color boxes, only the 5 options in red include grape.
Thus, P(grape in a 2-color box) = 5/15 = 1/3.