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Three six faces fair die are thrown together.What is the probability

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Three six faces fair die are thrown together.What is the probability  [#permalink]

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New post 24 Jul 2017, 11:25
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Three six faces fair die are thrown together.What is the probability that the sum of the numbers appearing on the die is 12?
(a) 27/216
(b) 25/216
(c) 23/216
(d) 17/216
(c) 24/216


A pretty easy problem, if one can find out the sum of numbers adding to 12 without listing. Any suggestions?

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Re: Three six faces fair die are thrown together.What is the probability  [#permalink]

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New post 24 Jul 2017, 11:35
theperfectgentleman wrote:
Three six faces fair die are thrown together.What is the probability that the sum of the numbers appearing on the die is 12?
(a) 27/216
(b) 25/216
(c) 23/216
(d) 17/216
(c) 24/216


A pretty easy problem, if one can find out the sum of numbers adding to 12 without listing. Any suggestions?

Total number of events = 6*6*6 = 216
To get the number 12 we will have to pair up the numbers on the die
Number = 1 on 1st die {156,165} Total 2
Number =2 on 1st die {246 , 264, 255} Total 3
Number = 3 on 1st die {363, 336, 354, 345} Total 4
Number = 4 on 1st die {462, 426, 453, 435, 444} Total 5
Number = 5 on 1st die {561, 516, 552, 525, 543, 534} Total 6
Number = 6 on 1st die {651, 615, 642, 624, 633} Total 5

So Number of events of getting 12 = 25

Probability = 25/216

Answer B
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Three six faces fair die are thrown together.What is the probability  [#permalink]

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New post 24 Jul 2017, 11:45
theperfectgentleman wrote:
Three six faces fair die are thrown together.What is the probability that the sum of the numbers appearing on the die is 12?
(a) 27/216
(b) 25/216
(c) 23/216
(d) 17/216
(c) 24/216

A pretty easy problem, if one can find out the sum of numbers adding to 12 without listing. Any suggestions?


With listing, possibly the easiest method.
For sum to be 12 with three dice, we could have a {(1,5,6),(2,4,6),(2,5,5),(3,3,6),(3,4,5),(4,4,4)}

For the combinations (1,5,6),(2,4,6) and (3,4,5) : There are 3! ways of arranging these numbers, \(3*3! = 18\)

For combinations (2,5,5),(3,3,6) : There are \(\frac{3!}{2!}\) ways of arranging these numbers, \(2*\frac{3!}{2!} = 6\)

There is only one way of arranging the (4,4,4) combination
There are a total of 25 favourable combinations.

The denominator for the probability is 6*6*6 = 216

Probability(sum appearing on three die is 12) : \(\frac{18+6+1}{216} = \frac{25}{216}\)(Option B)

--== Message from the GMAT Club Team ==--

THERE IS LIKELY A BETTER DISCUSSION OF THIS EXACT QUESTION.
This discussion does not meet community quality standards. It has been retired.


If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.

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Three six faces fair die are thrown together.What is the probability   [#permalink] 24 Jul 2017, 11:45
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