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Updated on: 20 Jun 2015, 03:33
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.
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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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
69% (01:59) correct
31%
(02:09)
wrong
based on 765
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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
(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-\(\frac{15}{6^4}\)
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-\(\frac{15}{6^4}\)
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 (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:
GMAT assassins aren't born, they're made,
Rich
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:
GMAT assassins aren't born, they're made,
Rich
20 Jun 2015, 03:21
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Mechmeera
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!
