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One has three fair dice and rolls them together. What is the

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New post 30 Sep 2003, 16:38
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One has three fair dice and rolls them together. What is the probability of having two same numbers and one different? (for example, 2-2-6, 1-3-3, 6-5-6, and so on)


1/6 *1 * 5/6 = 5/36 (1st and 2nd same, third different)

5/6 * 1 * 1/6 =5/36 (2nd and 3rd same, first different)

1/6 *5/6*1 = 5/36 (3rd and 1st same ,second different)


Total = 15/36 = 5/12

what say?

thanks
praetorian
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New post 03 Oct 2003, 05:31
Let's work this other way around:-

Total no. of outcomes : 6x6x6
P(No.of events with all 3 numbers same)=(1/216)x6=1/36
P(No. of events with all the numbers different on eah dice)=(6x5x4)/216=20/36

Total Probability= 21/36

Therefore, P(@ number same on outcome)= 1-21/36=15/36

VALIDATES YOUR ANSWER
  [#permalink] 03 Oct 2003, 05:31
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