Latest TCS Aptitude Question SOLUTION: Three dice are rolled. What is the probability of getting the sum as 13?

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(#M40040387) TCS Numerical Ability question Keep an EYE

Three dice are rolled. What is the probability of getting the sum as 13?

Asked In TCS Aakash Hegde (9 years ago)

Unsolved Read Solution (11)

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Read Solution (11) :

BHASKAR MANDAL  (9 years ago)

possibilities are
(1,6,6) -> 3!/2! = 3, as the same numbers can occur in different dices
(2,5,6) -> 3! = 6
(3,5,5) -> 3!/2! = 3
(3,4,6) -> 3! = 6
(4,4,5) -> 3!/2! = 3

total possibility = 3+6+3+6+3 = 21
total chances = 6^3 = 216
probability = 21/216 = 7/72 answer

Is this solution Helpfull? Yes (62) | No (5)  

hema  (9 years ago)

It is all u have to mug up the sequence of suming 3 dies:
to get sum of 3=1
4=3
5=6
6=10
7=15
8=21
9=25
10-27
similarly write the sequence in reverse order now,
11=27
12=25
13=21
14=15
15=10
16=6
17=3
18=1
so if u remember this then,few seconds enough to answer the prob of sum of 13
21/216

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apoorva trivedi  (9 years ago)

1/12 ass 6,6,1;5,6,2;4,6,3;3,6,4;2,6,5;1,6,6 for 1 dice.
like wise for other 2 dice.total is 6^3=216.
so 18/216 is 1/12.

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Palak Gupta  (9 years ago)

ans is 21/216

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PALLAVI MITRA  (9 years ago)

If the numbers of 2 dice are decided then only one number is possible on the third dieIf the value on the first die is 1 then the value on the 2nd die must be 6 then there is 1 possibility similarly if the value on 1st die is 2 then the value on 2nd die must be 5 or 6 similarly for all other combinationsTotal number of outcomes1345215Therefore probability15216total chances63

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Archana Gehlot  (9 years ago)

bhaskar is right 21/216

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chandra mondal  (9 years ago)

21/216=7/72(ANS)

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upasana acharya  (9 years ago)

the answr myt b 4

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tirupathi reddy  (9 years ago)

15
1,6,6 ----in 3 ways
2,5,6----- in 6 ways
3,4,6------in 6 ways

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Utkarsh Kumar Singh  (9 years ago)

2721618

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Supriya Mulpuru  (7 years ago)

possibilities are:
(6,6,1)
(6,5,2)
(6,4,3)
(6,3,4)
(6,2,5)
(6,1,6)
(5,6,2)
(5,5,3)
(5,4,4)
(5,3,5)
(5,2,6)
(4,6,3)
(4,5,4)
(4,4,5)
(4,3,6)
(3,6,4)
(3,5,5)
(3,4,6)
(2,6,4)
(2,4,6)
(1,6,6)

So if the sum starts with "6" we get 6 possibilities, similarly we 5 possibilities if started with 5, 4 with 4, 3 with 3, 2 with 2, 1 with 1.
sum of the possibilities is 6+5+4+3+2+1=21
total chances= 6^3 = 216
total possibilities= 21/216

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