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Probablity

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by Matt@VeritasPrep » Fri Aug 19, 2016 1:36 am
parveen110 wrote:
My approach:

6/6 * 1/6 * 5/6 * 4/6 = 5/54.

Where is this approach flawed?
You've essentially got:

First Die = Anything
Second Die = Same as first
Third Die = Different from first two
Fourth Die = Different from first three

This is a totally intuitive approach, but it's too restrictive: we don't have to be sure that the first two dice are the two that match. We also could have more than one pair, so we need to consider that scenario too.
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by YTarhouni » Sun Sep 03, 2017 1:24 pm
Use complement.
P(at least one pair)=1-P(no pairs).
P(no pair)=4C2 (possible combinations of pairs) *P(one number)*P(of not that one number)*P(not the two previous numbers)*P(not three previous numbers).
P(no pair)=6*1/6*5/6*4/6*3/6=60/216=5/18
1-5/18=13/18
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