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Re: A dog breeder currently has 9 breeding dogs. 6 of the dogs have exactl [#permalink]
sm021984 wrote:
A dog breeder currently has 9 breeding dogs. 6 of the dogs have exactly 1 littermate, and 3 of the dogs have exactly 2 littermates. If 2 dogs are selected at random, what is the probability that both selected dogs are NOT littermates?

A. 1/6
B. 2/9
C. 5/6
D. 7/9
E. 8/9


My thought process if it helps anyone:

out of 9 dogs we can pair them in 36 ways or \(9C2\)

Probability we can pick a sibling + probability we don't pick a sibling = 1

To pick a sibling we have 3 pairs and one triple

So:

\(2C2 + 2C2 + 2C2 + 3C2 = 6\)

there is \(\frac{1}{6}\) chance to select a sibling

Hence \(\frac{5}{6}\) we do not select a sibling.
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Re: A dog breeder currently has 9 breeding dogs. 6 of the dogs have exactl [#permalink]
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Re: A dog breeder currently has 9 breeding dogs. 6 of the dogs have exactl [#permalink]
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