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it is: consider the most expensive 2 books....than we have 6 slots free to combine the books. the total number of combs is 8C3 because we have a total of 8 books to combine in groups of three...prob=6/8*7=3/28

Re: (probability) 3 out of 8, the most expensive two included... [#permalink]

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09 Jan 2008, 06:44

i thought about it like this.

probability = desired outcomes / total out comes

where desired outcomes is the number of outcomes of picking 3 out of 8 books, but the two most expensive have to be in there, so ive got (1C1)*(1C1)*(6C1) / 8C3 = 6/56 = 3/28

I think 2! is permutation: \(P^2_2\). But you use \(C^8_3\) for all variants that mean ABC and BAC are the same variant. If we distinguish between ABC and BAC, we use \(P^8_3\) and \(P^3_3\)

Re: (probability) 3 out of 8, the most expensive two included... [#permalink]

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17 Feb 2010, 03:44

pretttyune wrote:

Please help me to understand this... I am always terrified when i face PROBABILITY questions...

Q. When choose 3 out of 8 books, the probability of the most expensive two books must be included?

Please helpe me kindly and thank you in advance...

Out of 8 .. 2 books (which are most exp) should be considered in your selection.

No of ways the 2 Exp books can be selected = 2c2 = 1

remaining we need to choose one more book from the 6 left over books = 6c1 = 6

Total ways of selecting 3 books out of 8 = 8c3

Prob = 1X6 / 8c3 = 3/28
_________________

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Re: (probability) 3 out of 8, the most expensive two included... [#permalink]

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05 Aug 2013, 21:14

Probability of selecting most expensive out of 8 =1/8 Probability of selecting next most expensive out of 7 =1/7 And out of 6 left out can select any one out of 6 = 6c1/6c1 now probability of selecting 2 most expensive = 1/8*1/7*1*3! =3/8(3! is because of arrangements)

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