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.
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.
