Combinations

[ellexay]

Example: Assume you have 20 M&M's® color distributed as above. If selected without replacement, in how many ways can you select two red ones in two selections? What is the corresponding probability?

Answer: For the first selection, five of the 20 M&M's® are red. Since we need to get two reds in only two selections, we need only consider this successful case further, ignoring what happens if we do not get a red on this first selection. For the second selection, only four red of the 19 M&M's® remain. Hence there are 5•4=20 ways of selecting two reds M&M's® in two selections.
The corresponding probability would be: (5/20)•(4/19)=20/380=1/19 or approximately 0.0526.

I understand the explanation. However, I would like to know why the following does not work:

5C1 * 4C1.

Thank you in advance.

[gaggleofgirls]

I am not sure why you say that 5c1 * 4c1 is not correct?

5c1 = 5
4c1 = 4

5*4 = 20, just like you already worked out.

-Carrie

Oh, and your question is difficult as it says "as above" when there is nothing above, so we don't know the distribution of M&M colors.

[Stuart@KaplanGMAT]

Example: Assume you have 20 M&M's® color distributed as above. If selected without replacement, in how many ways can you select two red ones in two selections? What is the corresponding probability?

Answer: For the first selection, five of the 20 M&M's® are red. Since we need to get two reds in only two selections, we need only consider this successful case further, ignoring what happens if we do not get a red on this first selection. For the second selection, only four red of the 19 M&M's® remain. Hence there are 5•4=20 ways of selecting two reds M&M's® in two selections.
The corresponding probability would be: (5/20)•(4/19)=20/380=1/19 or approximately 0.0526.

I understand the explanation. However, I would like to know why the following does not work:

5C1 * 4C1.

Thank you in advance.

Probability = # desired outcomes / total # of possibilities.

If you want to use combinatorics (which is a lot more work than the solution you posted), we'd so so as following:

There are 5 red M&Ms and we want to select 2 of them so there are 5C2 desired outcomes.

5C2 = 5*4/2*1 = 10 desired outcomes

There are 20 total M&Ms and we want to select 2 of them, so the total number of possibilities is 20C2.

20C2 = 20*19/2*1 = 380/2 = 190 total possibilities

Therefore, the probability of gettting 2 red is 10/190 = 1/19