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Old 2005 April 18th, 10:02 PM   #1 (permalink)
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probability
There are 8 books, 6 of them are paperbacks and 2 of them are hardbacks. If 4 books are to be selected at random, in how many ways, there is at least 1 hardback book?
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Old 2005 April 18th, 10:17 PM   #2 (permalink)
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Re: probability
6c3 * 2c1 + 6c2 * 2c2
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Old 2005 April 18th, 10:40 PM   #3 (permalink)
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Re: probability
6c3*2c1 + 6c2*1 = 55
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Old 2005 April 18th, 11:02 PM   #4 (permalink)
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Re: probability
8c4 - 6c4 = 55
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Old 2005 April 19th, 04:27 PM   #5 (permalink)
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Re: probability
Hey grey cellz. pl explain your response...

I get really confused with atleast in probability or combination Qs
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Old 2005 April 19th, 05:01 PM   #6 (permalink)
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Re: probability
8C4 is total ways in which you can select any 4 books out of 8
6C4 is the total ways you can select NO hardbacks
thus, total ways to select at least one hardback = 8C4 - 6C4
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Old 2005 April 27th, 09:06 PM   #7 (permalink)
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Re: probability
atleast one hardback book = 8C4 - No hardback book
8c4 - 6c4 = 70 - 15 =55
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