Problem Solving

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