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Urn 1 contains 4 red chips and 3 white chips. Urn 2 contains 3 red chips and 2 white chips. 2 chips are chosen at random and without replacement from each urn. 1) What is the probability that all 4 chips will be red? and 2) What is the probability of getting 3 red chips and 1 white chip amongst the 4?

Not homework, studying for a midterm and got very confused.

For 1) I got ((4 choose 2)(3 choose 2))/((7 choose 2)(5 choose 2))

For 2), in the numerator I think I need to add all the different mutually exclusive outcomes, so (this is just the numerator): (4c1)(3c1)(3c2) + (4c2)(3c1)(2c1), over the same denominator as above - (c indicates choose).

Is this correct? I always seem to have so much trouble with these counting type questions, like the poker hands, choosing objects from urns, rolling x die and choosing a specific number of 3s,4s etc... What's a general fool-proof, methodical way to approach these problems?

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You are right. You are right you need to be methodical to get all the cases and only once, but I don't know a simple mechanical one. – Ross Millikan Oct 12 '12 at 0:38
    
@RossMillikan Thanks. I guess practice makes perfect... – user44429 Oct 12 '12 at 0:52
1  
I can’t resist observing that if the urns are filled with chips, the probability of picking balls from them is zero. :-) – Brian M. Scott Oct 12 '12 at 8:27
up vote 0 down vote accepted

Answered in question description...

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