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If four fair dice are thrown simultaneously, what is the probability [#permalink]
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Updated on: 20 Jun 2015, 03:33
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If four fair dice are thrown simultaneously, what is the probability of getting at least one pair? (A) 1/6 (B) 5/18 (C) 1/2 (D) 2/3 (E) 13/18 One way to solve it is by considering the opposite  That on throwing the dice 4 times, you never get a pair.
In this case, the probability of what happens on the first throw does not matter. But the probability of getting a different number on the second, third and fourth throw are \(\frac{5}{6}, \frac{4}{6} and \frac{3}{6}\) respectively.
So on multiplying we get \(\frac{6}{6}*\frac{5}{6}*\frac{4}{6}*\frac{3}{6} = \frac{5}{18}\)
On subtracting it from 1, we get \(1(\frac{5}{18}) = \frac{13}{18}\)
Therefore E. \(\frac{13}{18}\) is the correct answer
But, I wonder, Why cant we solve by the following approach?
total possible outcomes for dice 6*6*6*6 or \(6^4\) considering the opposite outcomes or possible pairs =\(6C4=6C2=15\)
Thus, P(at least one pair)=1\(\frac{15}{6^4}\)
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Originally posted by Nevernevergiveup on 20 Jun 2015, 03:08.
Last edited by Bunuel on 20 Jun 2015, 03:33, edited 1 time in total.
Renamed the topic and edited the question.



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Re: If four fair dice are thrown simultaneously, what is the probability [#permalink]
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20 Jun 2015, 03:21
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Mechmeera wrote: If four fair dice are thrown simultaneously, what is the probability of getting at least one pair? \((A) \frac{1}{6} (B) \frac{5}{18} (C)\frac{1}{2} (D) \frac{2}{3} (E) \frac{13}{18}\)
One way to solve it is by considering the opposite  That on throwing the dice 4 times, you never get a pair.
In this case, the probability of what happens on the first throw does not matter. But the probability of getting a different number on the second, third and fourth throw are \(\frac{5}{6}, \frac{4}{6} and \frac{3}{6}\) respectively.
So on multiplying we get \(\frac{6}{6}*\frac{5}{6}*\frac{4}{6}*\frac{3}{6} = \frac{5}{18}\)
On subtracting it from 1, we get \(1(\frac{5}{18}) = \frac{13}{18}\)
Therefore E. \(\frac{13}{18}\) is the correct answer
But, I wonder, Why cant we solve by the following approach?
total possible outcomes for dice 6*6*6*6 or \(6^4\) considering the opposite outcomes or possible pairs =\(6C4=6C2=15\)
Thus, P(at least one pair)=1\(\frac{15}{6^4}\) Hi Mechmeera, You must Understand that the method of calculating total Outcomes = \(6^4\) includes arrangements of those outcomes on 4 dice But \(6C4=6C2=15\) does NOT include arrangements of outcomes on 4 dice Arrangement of 4 different outcomes (calculated by 6C4) on 4 dice can be calculated by = 4*3*2*1 = 4! so favorable outcomes in this case should be calculated as = \(6C4 * 4!=15 * 24 = 360\) Probability = Favorable Outcomes / Total CoutcomesOR Probability = 1 (Unfavorable Outocmes / Total Outcomesi.e. Probability = \(1 (360/6^4)\) = \(1 5/18 = 13/18\) I hope it clears your doubt!
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Re: If four fair dice are thrown simultaneously, what is the probability [#permalink]
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20 Jun 2015, 03:31
GMATinsight wrote: Mechmeera wrote: If four fair dice are thrown simultaneously, what is the probability of getting at least one pair? \((A) \frac{1}{6} (B) \frac{5}{18} (C)\frac{1}{2} (D) \frac{2}{3} (E) \frac{13}{18}\)
One way to solve it is by considering the opposite  That on throwing the dice 4 times, you never get a pair.
In this case, the probability of what happens on the first throw does not matter. But the probability of getting a different number on the second, third and fourth throw are \(\frac{5}{6}, \frac{4}{6} and \frac{3}{6}\) respectively.
So on multiplying we get \(\frac{6}{6}*\frac{5}{6}*\frac{4}{6}*\frac{3}{6} = \frac{5}{18}\)
On subtracting it from 1, we get \(1(\frac{5}{18}) = \frac{13}{18}\)
Therefore E. \(\frac{13}{18}\) is the correct answer
But, I wonder, Why cant we solve by the following approach?
total possible outcomes for dice 6*6*6*6 or \(6^4\) considering the opposite outcomes or possible pairs =\(6C4=6C2=15\)
Thus, P(at least one pair)=1\(\frac{15}{6^4}\) Hi Mechmeera, You must Understand that the method of calculating total Outcomes = \(6^4\) includes arrangements of those outcomes on 4 dice But \(6C4=6C2=15\) does NOT include arrangements of outcomes on 4 dice Arrangement of 4 different outcomes (calculated by 6C4) on 4 dice can be calculated by = 4*3*2*1 = 4! so favorable outcomes in this case should be calculated as = \(6C4 * 4!=15 * 24 = 360\) Probability = Favorable Outcomes / Total CoutcomesOR Probability = 1 (Unfavorable Outocmes / Total Outcomesi.e. Probability = \(1 (360/6^4)\) = \(1 5/18 = 13/18\) I hope it clears your doubt! Thank you very much for clearing out the confusion GMATinsight
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Re: If four fair dice are thrown simultaneously [#permalink]
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20 Jun 2015, 03:33
Mechmeera wrote: GMATinsight wrote: Mechmeera wrote: If four fair dice are thrown simultaneously, what is the probability of getting at least one pair? \((A) \frac{1}{6} (B) \frac{5}{18} (C)\frac{1}{2} (D) \frac{2}{3} (E) \frac{13}{18}\)
One way to solve it is by considering the opposite  That on throwing the dice 4 times, you never get a pair.
In this case, the probability of what happens on the first throw does not matter. But the probability of getting a different number on the second, third and fourth throw are \(\frac{5}{6}, \frac{4}{6} and \frac{3}{6}\) respectively.
So on multiplying we get \(\frac{6}{6}*\frac{5}{6}*\frac{4}{6}*\frac{3}{6} = \frac{5}{18}\)
On subtracting it from 1, we get \(1(\frac{5}{18}) = \frac{13}{18}\)
Therefore E. \(\frac{13}{18}\) is the correct answer
But, I wonder, Why cant we solve by the following approach?
total possible outcomes for dice 6*6*6*6 or \(6^4\) considering the opposite outcomes or possible pairs =\(6C4=6C2=15\)
Thus, P(at least one pair)=1\(\frac{15}{6^4}\) Hi Mechmeera, You must Understand that the method of calculating total Outcomes = \(6^4\) includes arrangements of those outcomes on 4 dice But \(6C4=6C2=15\) does NOT include arrangements of outcomes on 4 dice Arrangement of 4 different outcomes (calculated by 6C4) on 4 dice can be calculated by = 4*3*2*1 = 4! so favorable outcomes in this case should be calculated as = \(6C4 * 4!=15 * 24 = 360\) Probability = Favorable Outcomes / Total CoutcomesOR Probability = 1 (Unfavorable Outocmes / Total Outcomesi.e. Probability = \(1 (360/6^4)\) = \(1 5/18 = 13/18\) I hope it clears your doubt! Thank you very much for clearing out the confusion GMATinsightThe tradition of extending "Thank" on GMAT CLUB is pressing the +1KUDOS button Cheer!
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Re: If four fair dice are thrown simultaneously, what is the probability [#permalink]
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20 Jun 2015, 18:04
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Hi Mechmeera, In these types of probability questions, it's often easier to calculate what we DON'T WANT to have happen. We can then subtract that fraction from 1 to get the probability of what we DO want to have happen. Here, we're going to throw 4 (6sided) dice. We're asked for the probability of throwing AT LEAST one pair of matching numbers. Since it's actually easier to calculate how to throw 0 pairs of matching numbers (and then subtract that fraction from 1), I'll approach the question that way... 1st roll = any number = 6/6 2nd roll = not a match to the first = 5/6 3rd roll = not a match to the 1st or 2nd = 4/6 4th roll = not a match to the 1st or 2nd or 3rd = 3/6 (6/6)(5/6)(4/6)(3/6) = (1)(5/6)(2/3)(1/2) = 10/36 10/36 is the probability of rolling 0 matching numbers, so... 1  10/36 = 26/36 = the probability of rolling at least one matching pair of numbers 26/36 = 13/18 Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: If four fair dice are thrown simultaneously, what is the probability [#permalink]
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23 Oct 2016, 02:20
Proba (at least one pair) = 1  Proba (no pair)
Proba (no pair) = \(\frac{6}{6}*\frac{5}{6}*\frac{4}{6}*\frac{3}{6} = \frac{10}{36}\)
So, Proba (at least one pair) = 1  Proba (no pair) = \(1  \frac{10}{36} = \frac{26}{36} = \frac{13}{18}\)



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Re: If four fair dice are thrown simultaneously, what is the probability [#permalink]
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12 Nov 2016, 07:26
EMPOWERgmatRichC wrote: Hi Mechmeera, In these types of probability questions, it's often easier to calculate what we DON'T WANT to have happen. We can then subtract that fraction from 1 to get the probability of what we DO want to have happen. Here, we're going to throw 4 (6sided) dice. We're asked for the probability of throwing AT LEAST one pair of matching numbers. Since it's actually easier to calculate how to throw 0 pairs of matching numbers (and then subtract that fraction from 1), I'll approach the question that way... 1st roll = any number = 6/6 2nd roll = not a match to the first = 5/6 3rd roll = not a match to the 1st or 2nd = 4/6 4th roll = not a match to the 1st or 2nd or 3rd = 3/6 (6/6)(5/6)(4/6)(3/6) = (1)(5/6)(2/3)(1/2) = 10/36 10/36 is the probability of rolling 0 matching numbers, so... 1  10/36 = 26/36 = the probability of rolling at least one matching pair of numbers 26/36 = 13/18 Final Answer: GMAT assassins aren't born, they're made, Rich Hi Richard, Im still confused on how the 3 iteration yields a probability of (4/6) should it not be (5/6) and the 4th iteration yield another (5/6) Best Regards, Nik



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Re: If four fair dice are thrown simultaneously, what is the probability [#permalink]
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12 Nov 2016, 11:23
nikroshania1989 wrote: EMPOWERgmatRichC wrote: Hi Mechmeera, In these types of probability questions, it's often easier to calculate what we DON'T WANT to have happen. We can then subtract that fraction from 1 to get the probability of what we DO want to have happen. Here, we're going to throw 4 (6sided) dice. We're asked for the probability of throwing AT LEAST one pair of matching numbers. Since it's actually easier to calculate how to throw 0 pairs of matching numbers (and then subtract that fraction from 1), I'll approach the question that way... 1st roll = any number = 6/6 2nd roll = not a match to the first = 5/6 3rd roll = not a match to the 1st or 2nd = 4/6 4th roll = not a match to the 1st or 2nd or 3rd = 3/6 (6/6)(5/6)(4/6)(3/6) = (1)(5/6)(2/3)(1/2) = 10/36 10/36 is the probability of rolling 0 matching numbers, so... 1  10/36 = 26/36 = the probability of rolling at least one matching pair of numbers 26/36 = 13/18 Final Answer: GMAT assassins aren't born, they're made, Rich Hi Richard, Im still confused on how the 3 iteration yields a probability of (4/6) should it not be (5/6) and the 4th iteration yield another (5/6) Best Regards, Nik P(at least one pair)=1P(No pair) so to calculate p(No pair) Dice 1 6/6a Dice 2 5/6 (we got to exclude the number that has come on the face of Dice 1)b Dice 3 4/6 (we got to exclude the number that has come on the face of Dice 1 and on Dice 2; Remember, we don't want a pair)c Dice 43/6 (we got to exclude the number that has come on the face of Dice 1, Dice 2 and Dice 3; Remember, we don't want a pair)d p(No pair)=(a*b*c*d)=6/6*5/6*4/6*3/6=5/18 P(at least one pair)=1P(No pair)=>1(a*b*c*d)=15/18=13/18. +1 Kudos is your appreciation.
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Re: If four fair dice are thrown simultaneously, what is the probability [#permalink]
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12 Nov 2016, 12:04
Hi Nik, When you roll multiple dice, as you roll each die, you have to consider how the result will impact the overall calculation. In my explanation, I'm calculating the probability that NONE of the dice will match one another. The first die can be ANY of the 6 values. Once we've rolled that first die though, there are only 5 possibilities that don't match, so the probability is 5/6. Assuming that the second die does not match the first, we have two numbers that we're trying to 'avoid.' The third die would then have just 4 possibilities that don't match, so the probability is 4/6. Again, assuming that the third die does not match either of the first two, then we have three numbers to 'avoid.' With the 4th die, we only have 3 possibilities that don't match, so the probability is 3/6. GMAT assassins aren't born, they're made, Rich
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Re: If four fair dice are thrown simultaneously, what is the probability [#permalink]
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23 Apr 2018, 13:34
Nevernevergiveup wrote: If four fair dice are thrown simultaneously, what is the probability of getting at least one pair?
(A) 1/6 (B) 5/18 (C) 1/2 (D) 2/3 (E) 13/18
We want P(get at least 1 pair) When it comes to probability questions involving "at least," it's typically best to use the complement. That is, P(Event A happening) = 1  P(Event A not happening) So, here we get: P(getting at least 1 pair) = 1  P(not getting at least 1 pair)What does it mean to not get at least 1 pair? It means getting zero pairs. So, we can write: P(getting at least 1 pair) = 1  P(getting zero pairs)P(getting zero pairs)P(getting zero pairs)= P(ANY value on 1st die AND different value on 2nd die AND different value on 3rd die AND different value on 4th die) = P(ANY value on 1st die) x P(different value on 2nd die) x P(different value on 3rd die) x P(different value on 4th die) = 1 x 5/6 x 4/6 x 3/6 = 5/18 So, P(getting at least 1 pair) = 1  5/18 = 13/18 Answer: E Cheers, Brent
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Re: If four fair dice are thrown simultaneously, what is the probability
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