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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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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
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)
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