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April 28th, 2013, 03:16 PM   #1
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Can you help me figure out "at least one" probability questi

A cooler contains 11 cans of soda; 5 colas, 5 orange and 1 cherry. Two cans are selected at random without replacement. Find the probability that at least 1 can is cherry.

I need step-by-step help with this one, not just an answer. I have tried many examples and have no idea what I'm doing or even how to make sense of the example. Any help is greatly appreciated.
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April 28th, 2013, 04:13 PM   #2
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The chance that the first can chosen is cherry is 1/11.
The chance that the first can chosen is not cherry and the second can chosen is cherry is (10/11)(1/10) = 1/11.
Hence the required probability is 1/11 + 1/11 = 2/11.
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April 28th, 2013, 04:31 PM   #3
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Re: Can you help me figure out "at least one" probability qu

Hello, lilmscris!

Here is another approach.


Quote:
A cooler contains 11 cans of soda; 5 colas, 5 orange and 1 cherry.
Two cans are selected at random without replacement.
Find the probability that one can is cherry.









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April 28th, 2013, 04:39 PM   #4
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Re:

Quote:
Originally Posted by skipjack
The chance that the first can chosen is cherry is 1/11.
The chance that the first can chosen is not cherry and the second can chosen is cherry is (10/11)(1/10) = 1/11.
Hence the required probability is 1/11 + 1/11 = 2/11.
Thank you!!
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April 28th, 2013, 04:43 PM   #5
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Re: Can you help me figure out "at least one" probability qu

A third way: What are the chances of NOT getting a cherry in two draws? First draw, there is a 10/11 chance of not getting a cherry. Second draw, given the first draw, is 9/10 chance. (10/11)(9/10) = 9/11 chance of NOT drawing a cherry. So the chance of drawing a cherry is 1-(9/11) 0r 2/11

If you understand why all these methods gave the same result, you will understand the problem very well.
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April 28th, 2013, 05:02 PM   #6
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Re: Can you help me figure out "at least one" probability qu

Quote:
Originally Posted by johnr
A third way: What are the chances of NOT getting a cherry in two draws? First draw, there is a 10/11 chance of not getting a cherry. Second draw, given the first draw, is 9/10 chance. (10/11)(9/10) = 9/11 chance of NOT drawing a cherry. So the chance of drawing a cherry is 1-(9/11) 0r 2/11

If you understand why all these methods gave the same result, you will understand the problem very well.
This is a great explanation! Probably the one that made my brain hurt the least. Thank you so much!!
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