Three dices are thrown simultaneously. What is the probability of getting three numbers whose sum is

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04 Apr 2017 Probability

Three dices are thrown simultaneously. What is the probability of getting three numbers whose sum is 13?

OPtion

1) 14/216
2) 15/216
3) 16/216
4) 17/216
5) 18/216
6) 19/216
7) 10/116
8) 10/54
9) 5/54
10) None of these

Solution

Answer: 7/72
Total Number of possible outcomes = 6C1*6C1*6C1 = 216
Positive outcomes must satisfy D1+D2+D3 = 13
Positive outcomes which satisfy above condition:
1, 6, 6
2, 6, 5
2, 5, 6
3, 6, 4
3, 5, 5
3, 4, 6
4, 6, 3
4, 5, 4
4, 4, 5
4, 3, 6
5, 6, 2
5, 5, 3
5, 4, 4
5, 3, 5
5, 2, 6
6, 5, 2
6, 4, 3
6, 3, 4
6, 2, 5
6, 1, 6
6, 6, 1

Count of positive outcome = 21
Probability = Positive outcome/Total Possible Outcomes = 21/216 = 7/72
Option 10)

Correct Option: 10 Best Solution (6)

G.S.Kantharao   8 years AGO

Total outcomes 216
Required outcomes=(1,6,6),(2,5,6),(2,6,5),(3,4,6),(3,5,5),(3,6,4),(4,3,6),(4,4,5),(4,5,4),(4,6,3),(5,2,6),(5,3,5),(5,4,4),(5,5,3),(5,6,2),(6,1,6),(6,2,5),(6,3,4),(6,4,3),(6,5,2),(6,6,1)=21
Probability= 21/216= 7/72
Where the option is 10.

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Mukund Manohar   8 years AGO

Total Possible Outcome = 6^3 = 216
Positive Outcome = 21 [(1,6,6),(2,5,6),(2,6,5),(3,4,6),(3,5,5).(3,6,4),(4,3,6),(4,4,5),(4,5,4),(4,6,3),(5,2,6),(5,3,5),(5,4,4),(5,5,3),(5,6,2),(6,1,6),(6,2,5),(6,3,4),(6,4,3),(6,5,2) and (6,6,1)]
Required Probability = Positive Outcome/Total Possible Outcome
= 21/216 = 7/72

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Devendra Marghade   8 years AGO

Favorable outcomes are (1,6,6) (2,5,6) (2,6,5) (3,4,6) (3,5,5) (3,6,4) (4,3,6) (4,4,5) (4,5,4) (4,6,3) (5,2,6) (5,3,5) (5,4,4) (5,5,3) (5,6,2) (6,1,6) (6,2,5) (6,3,4) (6,4,3) (6,5,2) (6,6,1), total =21 and
Total outcomes=6x6x6=216
So probability=21/216 or 7/72
Option 10) None of these

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kandi sindhuja reddy   8 years AGO

ans : 9/216,
chances of getting sum =13 is (6,6,1)(6,1,6)(1,6,6)(6,5,2)(6,2,5)(2,6,5)(6,4,3)(3,6,4)(6,3,4)
total = 9
p(sum=13) = 9/216

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Abhinav   8 years AGO

Favourable outcomes = 21
Total outcomes - 6*6*6 = 216
P (sum is 13) = 21/216 =7/72

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palak Jain   8 years AGO

cases:
(1,6,6)---3
(2,6,5)----6
(3,6,4)----6
(5,5,3)----3
(4,4,5)------3
total=21
probability=21/216
=7/72

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