Problem Solving

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.

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.

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.)

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?

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

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.

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.

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

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.

__________

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