Latest TCS Aptitude Question SOLUTION: If three dies are thrown simultaneously probability of getting sum as 10?

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If three dies are thrown simultaneously probability of getting sum as 10?

Asked In TCS D Venkat Rami Reddy (10 years ago)

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m.s.l.n.pradeepthi  (10 years ago)

total possiblities =216
possibility to get sum 10 is (1,6,3)=3!=6 ways
(1,4,5)=3!=6ways
(2,6,2)=3!/2!=3ways
(2,5,3)=3!=6 ways
(2,4,4)=3!/2!=3ways
(3,3,4)=3!/2!=3ways
total probability=(3+3+6+6+3+6)/216=27/216

Is this solution Helpfull? Yes (42) | No (3)  

lavanya  (10 years ago)

Total number of posibilties =6^3=216
Getting sum 10=(1,4,5)-6ways
(1,3,6)-6ways
(2,4,4)-3ways
(2,3,5)-6ways
(2,2,6)-3ways
(3,3,4)-3ways
Total=21
Probability =21/216

Is this solution Helpfull? Yes (5) | No (13)  

SUBHOJIT DAS  (9 years ago)

Possible Sums with three dice are 3 ... 18

3 -> 1+1+1 Possible Combinations: 1

4 -> 1+2+1 Possible Combinations:3

5 -> (1+3+1),(1,2,2) Possible Combinations:6

6 -> (1,4,1), (1,3,2),(2,2,2) Possible Combinations:10

7 -> (1,4,2), (1,3,3),(5,1,1),(3,2,2) Possible Combinations:15

8 -> (1,4,3) , (1,2,5) ,(1,1,6),(4,2,2),(3,3,2) Possible Combinations:21

9 -> (6,2,1) , (5,3,1) ,(5,2,2), (4,4,1) ,(4,3,2),(3,3,3) Possible Combinations:25

10 -> (6,3,1) , (6,2,2) , (5,3,2), (5,4,1), (4,4,2),(4,3,3) Possible Combinations:27

11 -> (6,4,1),(6,3,2),(5,5,1),(5,4,2) (5,3,3),(4,4,3) Possible Combinations:27

12 -> (6,5,1),(6,4,2), (6,3,3),(5,5,2),(5,4,3),(4,4,4) Possible Combinations:25

13 -> (6,6,1) ,(6,5,2),(6,4,3),(5,5,3),(5,4,4) Possible Combinations:21

14 -> (6,4,4),(6,5,3),(5,5,4),(6,6,2) Possible Combinations::15

15-> (6,6,3),(6,4,5),(5,5,5) Possible Combinations:10

16-> (6,6,4),(6,5,5) Possible Combinations:6

17-> (6,6,5) Possible Combinations3

18-> 6,6,6 Possible Combinations:1

Henceforth answer 27/216

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Sunku Mounika  (10 years ago)

Probability of getting sum as 10 is 21/216.

Is this solution Helpfull? Yes | No (6)  

Ratan Dutta  (8 years ago)

3 dies=18 nos
we have to make a grp of 3 nos which sum=10...
we can permute 18 nos in different grp each consists 3=18P3
now the possible combinations are:-
1+6+3=10....6ways
2+2+6=10...3 ways
3+3+4....3 ways
2+5+3.....6 ways
4+4+2....3 ways
1+4+5....6 ways
=total 27 ways
probability of sum 10=27/18P3=3/544

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