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4 dices are thrown at the same time. What is the probability [#permalink]
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27 Dec 2009, 22:21
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4 dices are thrown at the same time. What is the probability of getting ONLY 2 dices showing the same face? A. 4/9 B. 5/9 C. 11/18 D. 7/9 E. none of the above what i have done is: total ways of getting values when 4 dices are thrown: 6^4 the 2 dices showing the same phase can be done in: 6 ways ............. (1) total ways of arranging these 4 face value: 4!/2!.................................(2) the remaing 2 places can be filled in 5c2 ways.....................................(3) [this is were the problem is......since the dices are thrown it should be arrangemt.. and when we use 5p2 the answer exceeds the value 6^4( not possible cause cause prob cant be m ore than 1) when 5c2 used: 5c2*4!/2!*6=720 when 5p2 used 1440>1296
and according mw it has to be permutation because 1 2 6 6 2 1 6 6 are different
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Last edited by Bunuel on 31 May 2013, 01:26, edited 2 times in total.
Edited the question and added the OA



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Re: Really tough [#permalink]
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4 dices are thrown at the same time. What is the probability [#permalink]
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28 Dec 2009, 10:08
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my take on this is.... choosing first no will have possibility of 1.... choosing the same on any of other three dice is \(\frac{1}{6}\) third dice can have any of remaining 5 so probability is \(\frac{5}{6.}\).. fourth dice can have any of remaining 4 so probability is \(\frac{4}{6}\)... so prob=\(1*\frac{1}{6}*\frac{5}{6}*\frac{4}{6}\)... .but there are six different values so it becomes =\(1*\frac{1}{6}*\frac{5}{6}*\frac{4}{6}*6=\frac{5}{9}\)..
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Re: Really tough [#permalink]
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07 Jan 2010, 01:43
Bunuel wrote: roshanaslam wrote: 4 dices are thrown at the same time. whats the probability of getting ONLY 2 dices showing the same face?? I suppose "only 2 dice showing the same face" means EXACTLY two? If so then: Total # of outcomes = 6^4 Favorable outcomes = 4C2=6, choosing two dice which will provide the same face, these two dice can take 6 values, other two 5 and 4. So, favorable outcomes=4C2*6*5*4. \(P=\frac{4C2*6*5*4}{6^4}=\frac{5}{9}\) Did the exact same thing. I hope that having similar thought processes to Bunuel's continues...
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Re: Really tough [#permalink]
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29 Sep 2011, 01:05
Please explain why am i going wrong  favourable outcomes  1st dice can have any number..hence, 6 possible ways 2nd dice ( same number ) = 1 way 3rd dice = 5 ways 4th dice = 4 ways total favourable outcome = 6*1*5*4 = 120 ways total outcome = 6*6*6*6 ways = 1296. Hence, probablity = 120/1296 = 5/54
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Re: Really tough [#permalink]
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29 Sep 2011, 05:28
krishnasty wrote: Please explain why am i going wrong 
favourable outcomes  1st dice can have any number..hence, 6 possible ways 2nd dice ( same number ) = 1 way 3rd dice = 5 ways 4th dice = 4 ways total favourable outcome = 6*1*5*4 = 120 ways
total outcome = 6*6*6*6 ways = 1296.
Hence, probablity = 120/1296 = 5/54 The problem is you assumed that 1st dice AND 2nd dice will be 6 and 1. That's not true. It could be: 2213 1223 1231 So, you must choose the index of the dices that will have the same sides. Favorable= 4C2*6*1*5*4
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Re: Really tough [#permalink]
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29 Sep 2011, 06:19
You are missing the number of permutations. 6*1*5*4 is perfect. But take an example the four digits are 1,1,3,4. This can further be arranged in 4!/2! ways. Hope that clarifies!!



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Re: Really tough [#permalink]
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29 Sep 2011, 06:41
munish4130 wrote: You are missing the number of permutations. 6*1*5*4 is perfect. But take an example the four digits are 1,1,3,4. This can further be arranged in 4!/2! ways. Hope that clarifies!! Actually combination... 4C2 = 4! / 2! (42)!



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Re: Really tough [#permalink]
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19 Oct 2011, 20:13
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chetan2u wrote: my take on this is.... choosing first no will have possibility of 1.... choosing the same on any of other three dice is 1/6 . third dice can have any of remaining 5 so probab is 5/6... fourth dice can have any of remaining 4 so probab is 4/6... so prob=1*1/6*5/6*4/6....but there are six different values so it becomes =1*1/6*5/6*4/6*6=5/9.. anyone if i have gone wrong Done in the same manner: Probability of getting two same faces: 1 and 1/6 For other two faces, probabilities: 5/6 and 4/6 respectively. The dice combinations can be arranged in 4!/2! ways Total probability = 1*1/6*5/6*4/6*4!/2! = 5/9



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Re: Really tough [#permalink]
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04 Nov 2011, 05:33
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\(= (\frac{1}{6}*\frac{1}{6}*6)*\frac{5}{6}*\frac{4}{6}*\frac{4!}{2!*2!}\) \(= \frac{5}{9}\) we multiply by 6 because there are 6 possibilities of having same numbers {1,1}.. {6,6} etc. we multiply by 4! for possible combinations in between the four dices, by 2! for two same numbers and again by 2! for possibilities between same & notsame numbers.
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Re: Really tough [#permalink]
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17 Dec 2011, 20:52
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Here is my approach:
If we name the dices: 1234 there are 6 possibilities that just 2 of them are the same: \(1&21&31&42&32&43&4\) and to calculate probability of the first one, 1&2, we know that the first dice can be every thing. but the second should be the same as first. the third should be something else than the first and second, and finally the forth should be something else than the three previous ones. So: \(1*\frac{1}{6}*\frac{5}{6}*\frac{4}{6}\)
and as i said, we have 6 possibilities. So total probability is: \(6*1*\frac{1}{6}*\frac{5}{6}*\frac{4}{6}=\frac{5}{9}\)



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Re: Really tough [#permalink]
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19 Dec 2011, 11:47
Say: only 1st and 2nd show the same face possible outcome 6^4
1st dice = 6c1=6 2nd dice= 1c1 [same as 1st]=1 this can happen in 6 different ways
3rd dice= 5c1=5 4th dice= 4c1=4
(6*1*6*5*4)/6^4=5/9



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Re: Really tough [#permalink]
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03 Jul 2012, 22:52
BDSunDevil wrote: Say: only 1st and 2nd show the same face possible outcome 6^4
1st dice = 6c1=6 2nd dice= 1c1 [same as 1st]=1 this can happen in 6 different ways
3rd dice= 5c1=5 4th dice= 4c1=4
(6*1*6*5*4)/6^4=5/9 I also did it this way. Now, the only difference btw Bunuel's method and this one is that we use 6C1 instead of 4C2. Can anybody help me understand if this approach is wrong in any way?
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Re: Really tough [#permalink]
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Re: Really tough [#permalink]
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04 Jul 2012, 01:58
Bunuel wrote: Aki wrote: BDSunDevil wrote: Say: only 1st and 2nd show the same face possible outcome 6^4
1st dice = 6c1=6 2nd dice= 1c1 [same as 1st]=1 this can happen in 6 different ways
3rd dice= 5c1=5 4th dice= 4c1=4
(6*1*6*5*4)/6^4=5/9 I also did it this way. Now, the only difference btw Bunuel's method and this one is that we use 6C1 instead of 4C2. Can anybody help me understand if this approach is wrong in any way? It's not clear what does the red part in the above solution mean. If it means that there are total of 4C2=6 ways to chose which two show the same face then the logic is correct. Well I'm not sure about BDSunDevil, but for me the red part means : there are 6 different ways the 1st and 2nd dice can correlate. i.e if both D1 and D2 :1 OR if both D1 and D2 :2 and so on till 6. Let me further elaborate: 1. We know that the both die can have one of the following numbers i.e 16. So possible combinations for D1 and D2: 6C1*1C1 (since D1 and D2 will have to be the same number) 2. Now, the 3rd dice can be any of the remaining 5 numbers. So, possible combinations: 5C1 3. Finally, the 4th dice can be any of the remaining 4 numbers. So, possible combinations: 4C1 From 1,2 and 3: Total Favorable Combinations: 6C1*1C1*5C1*4C1 Total Outcomes(as per your logic): 6^4 So, P(F)= (6C1*1C1*5C1*4C1)/ (6^4) I have no problems with your logic either, just want to know if this one is possible also.
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Probability  This is for Bunuel [#permalink]
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21 Jul 2012, 15:14
4 dices are thrown at the same time. whats the probability of getting ONLY 2 dices showing the same face? I suppose "only 2 dice showing the same face" means EXACTLY two? If so then:
Total # of outcomes = 6^4
Favorable outcomes = 4C2=6, choosing two dice which will provide the same face, these two dice can take 6 values, other two 5 and 4. So, favorable outcomes=4C2*6*5*4.
Answer: 5/9.
************
This was a solution posted by you on probability can you please explain why is the 6 shown twice ??
why is the 6 given twce it shuld be only ONCE
becuase the only possibility of 2 dices getting the same number is
66 55 44 33 22 11
that is 6 possibilities
so 6*5*4/6^4
why is the 6C2 in the picture
please explain



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Re: Probability  This is for Bunuel [#permalink]
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22 Jul 2012, 03:01
venmic wrote: 4 dices are thrown at the same time. whats the probability of getting ONLY 2 dices showing the same face? I suppose "only 2 dice showing the same face" means EXACTLY two? If so then:
Total # of outcomes = 6^4
Favorable outcomes = 4C2=6, choosing two dice which will provide the same face, these two dice can take 6 values, other two 5 and 4. So, favorable outcomes=4C2*6*5*4.
Answer: 5/9.
************
This was a solution posted by you on probability can you please explain why is the 6 shown twice ??
why is the 6 given twce it shuld be only ONCE
becuase the only possibility of 2 dices getting the same number is
66 55 44 33 22 11
that is 6 possibilities
so 6*5*4/6^4
why is the 6C2 in the picture
please explain We have 4 dice (A, B, C, D), not 2. So, \(C^2_4\) is the ways to select which two dice out of 4, will provide the same face: (A, B), (A, C), (A, D), (B, C), (B, D), or (C, D) . Next, these two dice can take 6 values, the remaining two 5 and 4, respectively. So, the # of favorable outcomes is \(C^2_4*6*5*4\). Hope it's clear.
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Re: Really tough [#permalink]
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27 Feb 2013, 16:54
Aki wrote: BDSunDevil wrote: Say: only 1st and 2nd show the same face possible outcome 6^4
1st dice = 6c1=6 2nd dice= 1c1 [same as 1st]=1 this can happen in 6 different ways
3rd dice= 5c1=5 4th dice= 4c1=4
(6*1*6*5*4)/6^4=5/9 I also did it this way. Now, the only difference btw Bunuel's method and this one is that we use 6C1 instead of 4C2. Can anybody help me understand if this approach is wrong in any way? 6C1 refers to the number of possibilities for the two same faced dice can have (1, 2, 3, 4, 5, 6). This is wrong as it was already counted for in the calculation. 4C2 refers to the number of possibilities the four dice can be arranged (1/1/2/3 as opposed to 2/3/1/1). However, I think the question was not clear as to whether the order of the four dice matters...



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Re: Really tough [#permalink]
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30 May 2013, 20:31
shahideh wrote: Here is my approach:
If we name the dices: 1234 there are 6 possibilities that just 2 of them are the same: \(1&21&31&42&32&43&4\) and to calculate probability of the first one, 1&2, we know that the first dice can be every thing. but the second should be the same as first. the third should be something else than the first and second, and finally the forth should be something else than the three previous ones. So: \(1*\frac{1}{6}*\frac{5}{6}*\frac{4}{6}\)
and as i said, we have 6 possibilities. So total probability is: \(6*1*\frac{1}{6}*\frac{5}{6}*\frac{4}{6}=\frac{5}{9}\) Is it just a coincidence that Bunuel's method, with 4C2 = 6, is the same as the # of pairings you had (1&21&3...etc). I don't understand why those should be the same #. 4C2 is simply the combinations of 2 dice chosen out of 4. Is yours a long form way of saying the same thing?



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Re: 4 dices are thrown at the same time. whats the probability [#permalink]
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31 May 2013, 01:28
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1. The formula is\(nCr\) 2. n is 6 as a dice has 6 values 3. r is 1 as only one value is shown 4. But there are 4 dices and there is a constraint on the value of n for 3 dices 5. Let the first dice show any value. Thus it can show 6 values and n is 6. r is 1 6. The second dice in consideration has to show the same value. Therefore n is 1. r is 1 7.The third dice cannot show the value on the first two dices and therefore n is 5. r is 1 8. Similarly for the fourth dice n is 4. r is 1 9. We can select 2 dices out of 4 dices in \(4C2\)ways 10. The total number of ways of exactly 2 dices out of 4 dices showing the same face is \(4C2 *6C1*1C1*5C1*4C1= 720\) 11. The total number of possibilities with any value with 4 dices is 6^4=1296 12. The probability is 720/1296 = 5/9
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