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.
Combinations
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- gaggleofgirls
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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.
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
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Probability = # desired outcomes / total # of possibilities.ellexay wrote: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.
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
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