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    myrmiller's Avatar
    myrmiller Posts: 2, Reputation: 1
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    #1

    Sep 8, 2011, 04:52 PM
    Probability of at least 1 orange can without replacement
    How do I calculate if I have 8 cans of soda: 2 colas, 5 orange, 1 cherry. 2 cans are selected at random without replacement. What is probability that at least one can is orange
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #2

    Sep 9, 2011, 09:19 AM
    When you have questions with 'at least', you can do 1 - P(none).

    Here, the P(none) will be P(no orange cans are picked).

    What is the probability that out of the 2 cans randomly picked, none are orange?

    Then, subtract that value from one.

    Can you post your answer? :)
    myrmiller's Avatar
    myrmiller Posts: 2, Reputation: 1
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    #3

    Sep 9, 2011, 09:39 AM
    Out of 8 cans 3/8 are non-orange. I think that probability of 2 cans not being orange is 3/8 X 2/7
    or 6/56. Subtracted from one is 50/56, or 25/28. That is the answer. THANK YOU!
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #4

    Sep 9, 2011, 09:49 AM
    Indeed! Well done myrmiller! :)
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