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 April 28th, 2013, 03:16 PM #1 Newbie   Joined: Apr 2013 Posts: 3 Thanks: 0 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.
 April 28th, 2013, 04:13 PM #2 Global Moderator   Joined: Dec 2006 Posts: 20,969 Thanks: 2219 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.
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.

$\text{There are: }\:{11\choose2} \:=\:\frac{11!}{2!\,9!} \:=\:55\text{ possible outcomes.}$

$\text{There are: }\:{10\choose2} \:=\:\frac{10!}{2!\,8!} \:=\:45\text{ ways to get }no\text{ cherry.}$

$\text{Hence, there are: }\:55\,-\,45\:=\:10\text{ ways to get one cherry.}$

$\text{Therefore: }\:P(\text{one can is cherry}) \:=\:\frac{10}{55} \:=\:\frac{2}{11}$

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!!

 April 28th, 2013, 04:43 PM #5 Math Team   Joined: Apr 2012 Posts: 1,579 Thanks: 22 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.
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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# a cooler contains 5 cans of soda 2 cola 2 orange 1 cherry. two cans are selected at random without replacement. find the probability that at least one can is cola

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