What is the probability of getting a sum 12 from three throws of a dice?

M4maths Previous Puzzles

19 Dec 2016 Probability

What is the probability of getting a sum 12 from three throws of a dice?

OPtion

1) 11/108
2) 7/72
3) 1/9
4) 25/216
5) 13/108
6) 1/8
7) 7/54
8) 29/216
9) 5/36
10)31/216

Solution

Total Number of possible outcomes = 6C1*6C1*6C1 = 6*6*6 = 216
Positive outcomes must satisfy D1+D2+D3= 12
(1,5,6) (1,6,5) (2,4,6) (2,6,4) (2,5,5) (3,3,6) (3,6,3) (3,4,5) (3,5,4) (4,2,6) (4,6,2) (4,3,5) (4,5,3) (4,4,4) (5,1,6) (5,6,1) (5,2,5) (5,5,2) (5,3,4) (5,4,3) (6,1,5) (6,5,1) (6,2,4) (6,4,2) (6,3,3)
Count of positive outcome = 25
Probability = Positive outcome/Total Possible Outcomes = 25/216

Option 4)

Correct Option: 4 Best Solution (18)

Mukund Manohar   8 years AGO

Sum of 12 can be achieved in following ways
6,5,1 Total cases = 3! = 6
6,4,2 Total cases = 3! = 6
6,3,3 Total cases = 3!/2! = 3
5,5,2 Total cases = 3!/2! = 3
5,4,3 Total cases = 3! = 6
4,4,4 Total cases = 3!/3! = 1
Total cases = 25
Required Probability = 25 * (1/6 * 1/6 * 1/6) = 25/216

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Jagriti Priya   8 years AGO

sum = 12
6,5,1 total case = 3! =6
6,4,2 total case = 3! =6
6,3,3 total case = 3!/2! = 3
5,5,2 total case = 3!/2! = 3
5,4,3 total case = 3! = 6
4,4,4 total case = 1! = 1
total cases = 25
probability = 25 *(1/6 * 1/6 * 1/6) = 25/216

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Pavan kumar   8 years AGO

Sum=12 is formed by following combinations n(a)= {(1,5,6),(1,6,5),(2,4,6),(2,5,5),(2,6,4),(3,3,6),(3,4,5),(3,5,4),(3,6,3),(4,2,6),(4,3,5),(4,4,4),(4,5,3),(4,6,2),(5,1,6),(5,2,5),(5,3,4),(5,4,3),(5,5,2),(5,6,1),(6,1,5),(6,2,4),(6,3,3),(6,4,2),(6,5,1)
N(s)= 6^3=216
P(a)=n(a)/n(s)
=25/216

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ARYASREE   8 years AGO

total no. of outcomes=6*6*6=216
cases getting sum 12=
(1,5,6),(1,6,5),(2,4,6),(2,5,5)(2,6,4),(3,3,6),(3,4,5),(3,5,4),(3,6,3),(4,2,6)(4,3,5,),(4,4,4),(4,5,3)(4,6,2),(5,1,6),(5,2,5),(5,3,4),(5,4,3),(5,5,2)(5,6,1),(6,1,5)(6,2,4),(6,3,3),(6,4,2),(6,5,1)
so no.favourable outcome that is getting sum of 12=25
so probability=no. of favourable outcome/total outcome
=25/216

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madanpal vishwakarma    8 years AGO

For three dice total outcome is 6^3= 216 and the combination to get sum 12 is
1) 4+3+5
2)4+4+4
3)5+2+5
4)6+4+2
5)6+3+3

As per above the number of favorable outcome = 6+1+6+6+6 =25

so the probability of getting sum 12 is = 25/216

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Dr.Dipin Singh   8 years AGO

Total possible combinations are 6*6*6=216.
Required combinations are
6+5+1... 6 combinations
6+4+2... 6 combinations
6+3+3... 3 combinations
5+4+3... 6 combinations
5+5+2 .. 3 combinations
4+4+4 .. 1 Combination

Total 25 favourable combinations.

so probability = 25/216.. option 4

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Pargat Pal Singh   8 years AGO

out of 216 possibilities, only 25 possibilities give sum of 12.
so prob=25/216

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G.S.Kantharao   8 years AGO

Total No. of out comes = 6³
No. of feasible of comes=25
(1,5,6),(2,4,6),(3,3,6),(5,1,6),(4,2,6),(1,6,5),(2,5,5),(3,4,5),(6,1,5),(5,2,5),(4,3,5),(2,6,4),(3,5,4),(4,4,4),(6,2,4),(5,3,4),(3,6,3)(4,5,3),(5,4,3),(6,3,3),(4,6,2),(6,4,2),(5,5,2),(5,6,1),(6,5,1) = 25
Hence the Probability = (25/216)
Right option is 4.

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santosh kumar   8 years AGO

6,5,1= 3!
6,4,2= 3!
6,3,3=3!/2!
5,5,2=3!/2!
5,4,3=3!
4,4,4=1!
total cases=25
probability= 25/(6*6*6)
= 25/216

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Priya Mahanta   8 years AGO

(6,5,1)= 3!
(4,4,4)= 1!
(3,4,5)= 3!
(2,6,4)= 3!
(2,5,5)= 3 ways
(3,3,6) = 3 ways
P(12)= 25/216

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Lakshmi Narayanan.k   8 years AGO

1,(5,6)=2
2,(4,5,6)=3
3,(3,4,5,6)=4
4,(2,3,4,5,6)=5
5,(1,2,3,4,5,6)=6
6,(1,2,3,4,5)=5
2+3+4+5+6+5=25
Probability= 25/216 (opt 4)

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souvik dutta   8 years AGO

25 combinations are possible(FOR MAKING 12) having 6*6*6=216 as sample space. hence answer is 25/216.

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Abhishek Gupta   8 years AGO

Sum of 12 can be achieved in following ways
6,5,1---Total cases = 3! = 6
6,4,2---Total cases = 3! = 6
6,3,3---Total cases = 3!/2! = 3
5,5,2---Total cases = 3!/2! = 3
5,4,3---Total cases = 3! = 6
4,4,4---Total cases = 3!/3! = 1
Total cases = 25
Probability = 25/(6*6*6) = 25/216

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pooja patil   8 years AGO

the sum 12 occurs in the following throws
(1,5,6)
(2,4,6)
(3,4,5)
each trial can occur in 6 possible ways like(1,5,6),(1,6,5),(6,1,5),(6,5,1),(5,6,1),(51,6)
so for the above trials there are 18 possible ways
(2,5,5),(5,2,2)(5,2,5)
(3,3,6),(6,3,6),(6,3,3)
(4,4,4)
P(E)=n(E)/n(S)
n(S)=216
n(E)=25
p(E)=25/216

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Somesh   8 years AGO

because in total 25 cases are possible in which sum can be equal to 12 out of 216

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RANJEET KUMAR   8 years AGO

6,5,1
.
.
.
4,4,4 in such way n(E)= 25
and n(S)= 216
hence p(E)= 25/216

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Vivek   8 years AGO

6,5,1 --> 3! = 6
6,4,2 --> 3! = 6
6,3,3 --> 3!/2! = 3
5,5,2 --> 3!/2! = 3
5,4,3 --> 3! = 6
4,4,4 --> 3!/3! = 1
Total = 25
Probability = 25 * (1/6 * 1/6 * 1/6) = 25/216

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Arshi   8 years AGO

1,5,6=3!
2,5,5=3!/2!
3,5,4=3!
4,4,4=3!/3!
3,6,3=3!/2!
6,4,2=3!/2!
25*(1/6*1/6*1/6)=25/216

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