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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?
(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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24 Jul 2017, 11:35
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theperfectgentleman
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
24 Jul 2017, 11:45
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theperfectgentleman
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)
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
