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 July 28th, 2013, 08:38 AM #1 Member   Joined: Sep 2011 Posts: 99 Thanks: 1 Probability doubts part 2 Here, I have another qns. I have gotten( 1/6*5/6*6/6) + (6/6*5/6*1/6) + (6/6*1/6*1/6) = 11/36 for a(ii). I am not sure if my answer is correct or not as i find that my answer seems logical from the approach used in the previous doubts stated above. Thank you in advance. Three unbiased dice are thrown. Find the probability that they (a) (i) all show different numbers (ii) at least two show the same numbers. Answer for a(i) is 6/6 * 5/6 * 4/6 = 5/9 a(ii) is ( 6/6*1/6*1/6) + (6/6*1/6*5/6) = 1/
 July 28th, 2013, 10:45 AM #2 Senior Member     Joined: Jul 2012 From: DFW Area Posts: 635 Thanks: 96 Math Focus: Electrical Engineering Applications Re: Probability doubts part 2 Hi [color=#0000BF]Alexis87[/color], For a(ii), I get: $3 \cdot \frac{6}{6} \cdot \frac{1}{6} \cdot \frac{5}{6}+\frac{6}{6} \cdot \frac{1}{6} \cdot \frac{1}{6}=3 \cdot \frac{5}{36}+\frac{1}{36}=\frac{15+1}{36}=\frac{16 }{36}=\frac{4}{9}$ The product of fractions in the first term on the left hand side is the probability that two individual die have the same value and the other individual die has a different value. If the dice are arbitrarily numbered 1 through 3, there are 3 ways that this can occur: 1) Die 1 and 2 are equal, 3 different; 2) Die 1 and 3 are equal, 2 different; 3) Die 2 and 3 are equal, 1 different. Therefore the individual probability is multiplied by 3. The last term on the left hand side is the probability that all are the same. I think that you just left out one of the possibilities, otherwise you were on the right track. Notice that a(i) and a(ii) include all possibilities for rolling the 3 dice, therefore the probabilities must add to 1.
 July 28th, 2013, 11:01 AM #3 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116 Re: Probability doubts part 2 Hi jks, So, you could have just subtracted 5/9 from 1 to get the answer to (ii) without any further calculations!
 July 28th, 2013, 11:10 AM #4 Senior Member     Joined: Jul 2012 From: DFW Area Posts: 635 Thanks: 96 Math Focus: Electrical Engineering Applications Re: Probability doubts part 2 Yes, indeed. But unless it is too calculation intensive, I like to work the problems both ways as a check.
 July 28th, 2013, 11:29 AM #5 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116 Re: Probability doubts part 2 It could also go like this: The prob that the first two are the same = 1/6. If the first two are different (Prob = 5/6), then the third is the same as one of these with prob = 2/6. This gives 1/6 + (5/6)(2/6) = 16/36 = 4/9 I guess that settles it!

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