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Probability

Expert replies
by 2008nv » Mon Mar 10, 2008 12:46 pm
A cupboard holds 10 cans, of which 3 contain pumpkin puree. If a group of 4 cans is randomly selected from the cupboard , what is the probability that the group includes the 3 cans containing pumpkin puree?

1/120
1/30 (correct)
1/21
3/10
4/10

I have 1/120
3/10*2/9*1/8*7/7
explain please :?:
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Source: — Problem Solving |

by rey.fernandez » Mon Mar 10, 2008 1:34 pm
You're almost there. Your calculation assumes that you're picking the 3 cans of pumpkin puree first, and then one of the other cans. But that's not the only way it could happen. Here are all the options: (P = pumpkin puree, N = not pumpkin puree)

P, P, P, N
P, P, N, P
P, N, P, P
N, P, P, P

There are four ways, each one having a probability of 1/120 as you calculated. So, 4 * 1/120 = 1/30.
Rey Fernandez
Instructor
Manhattan GMAT
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by 2008nv » Mon Mar 10, 2008 1:43 pm
Thank you :wink:
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by soumitb » Wed Mar 12, 2008 4:46 pm
This can also be done by the combination formula:

1. Total outcomes:
How many ways can we select 4 cans from 10 cans?
10C4 = 210

2. Desired outcomes (We need to select 4 cans):
1) Select 3 pumpkin puree cans out of 3 pumpkin puree cans
3C3 = 1

2) Select 1 "Non pumpkin" can out of 7
7C1 = 7

Therefore desired outcomes = 3C3 x 7C1 = 1 X 7 = 7

Therefore probability = (Desired outcomes) / (Total outcomes)
= 7/210
= 1/30
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by Jeff@TargetTestPrep » Thu Dec 14, 2017 7:43 am
2008nv wrote:A cupboard holds 10 cans, of which 3 contain pumpkin puree. If a group of 4 cans is randomly selected from the cupboard , what is the probability that the group includes the 3 cans containing pumpkin puree?

1/120
1/30
1/21
3/10
4/10

We have a total of 10 cans, 3 of which contain pumpkin puree. We need to determine the probability that a group of 4 randomly selected cans will contain the 3 pumpkin puree cans.

The number of ways to select 3 pumpkin puree cans from 3 is 3C3 = 1. The number of ways to select 1 can of non-pumpkin puree is 7C1 = 7. Thus, the number of ways to select 3 pumpkin puree cans and 1 non-pumpkin puree can is 1 x 7 = 7.

The total number of ways to select 4 cans from 10 is:

10C4 = 10!/[4!(10-4)!] = 10!/(4!6!) = (10 x 9 x 8 x 7)/4! = (10 x 9 x 8 x 7)/(4 x 3 x 2) = 10 x 3 x 7 = 210

Thus, the probability is 7/210 = 1/30.

Alternate Solution:

Let’s first determine the probability of choosing P-P-P-N, in which P stands for a pumpkin puree can choice and N stands for a non-pumpkin puree choice:

In the first draw, there are 3 pumpkin puree cans out of a total of 10 cans, so the probability of selecting a pumpkin puree can is 3/10.

In the second draw, there are 2 pumpkin puree cans out of a total of 9 cans, so the probability of selecting a pumpkin puree can is 2/9.

In the third draw, there is 1 pumpkin puree can out of a total of 8 cans, so the probability of selecting a pumpkin puree can is 1/8.

In the last draw, there is no pumpkin puree can left, so the probability of selecting a non-pumpkin puree can is 100%, or 1.

Thus, the probability of P-P-P-N is 3/10 x 2/9 x 1/8 x 1 = 1/10 x 1/3 x 1/4 = 1/120.

Now, we must observe that getting 3 pumpkin puree cans in a total of 4 draws can happen in 4 ways (namely P-P-P-N, P-P-N-P, P-N-P-P, and N-P-P-P), all with an equal chance of happening.

Thus, the probability of choosing 3 pumpkin puree cans in a total of 4 draws is 4 x 1/120 = 1/30.

Answer: B

Jeffrey Miller
Head of GMAT Instruction
[email protected]

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