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If four fair dice are thrown simultaneously, what is the probability

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If four fair dice are thrown simultaneously, what is the probability [#permalink]

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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

[Reveal] Spoiler:
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-\(\frac{15}{6^4}\)
[Reveal] Spoiler: OA

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Last edited by Bunuel on 20 Jun 2015, 02:33, edited 1 time in total.
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Re: If four fair dice are thrown simultaneously, what is the probability [#permalink]

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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 Coutcomes

OR Probability = 1- (Unfavorable Outocmes / Total Outcomes

i.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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New post 20 Jun 2015, 02:31
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 Coutcomes

OR Probability = 1- (Unfavorable Outocmes / Total Outcomes

i.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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New post 20 Jun 2015, 02:33
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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 Coutcomes

OR Probability = 1- (Unfavorable Outocmes / Total Outcomes

i.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, what is the probability [#permalink]

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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 (6-sided) 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:
[Reveal] Spoiler:
E


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Re: If four fair dice are thrown simultaneously, what is the probability [#permalink]

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New post 23 Oct 2016, 01:20
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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New post 12 Nov 2016, 06:26
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 (6-sided) 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:
[Reveal] Spoiler:
E


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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New post 12 Nov 2016, 10:23
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 (6-sided) 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:
[Reveal] Spoiler:
E


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)=1-P(No pair)
so to calculate p(No pair)
Dice 1- 6/6----a
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 4-3/6 (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)=1-P(No pair)=>1-(a*b*c*d)=1-5/18=13/18.

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Re: If four fair dice are thrown simultaneously, what is the probability [#permalink]

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New post 12 Nov 2016, 11:04
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.

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Re: If four fair dice are thrown simultaneously, what is the probability [#permalink]

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Re: If four fair dice are thrown simultaneously, what is the probability   [#permalink] 29 Nov 2017, 18:54
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