Latest TCS Aptitude Question SOLUTION: if 3 dices are rolled simultaneously then find probability that sum is 10?

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(#M40026055) TCS Numerical Ability question prob Keep an EYE

if 3 dices are rolled simultaneously then find probability that sum is 10?

Asked In TCS hitesh kapila (11 years ago)

Unsolved Read Solution (9)

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Ratikant Panda  (11 years ago)

1/8
as 10 can be obtained as (1,6,3)->6 ways;(1,4,5)->6 ways;(2,4,4)->3 ways;
(2,6,2)->3 ways;(2,5,3)->6 ways;(3,3,4)->3 ways.
total ways=27
probability=27/216=1/8

Is this solution Helpfull? Yes (19) | No (2)  

Himank Agrawal  (11 years ago)

Possible ways :-
(6,3,1) - 6
(6,2,2) - 3
(5,3,2) - 6
(5,4,1) - 6
(4,4,2) - 3
(4,5,1) - 6
(4,3,3) - 3
Total no of possibilites = 33
probability = 33/216 = 11/72

Is this solution Helpfull? Yes (8) | No (15)  

Ann Mary Jolly  (11 years ago)

Ans : 0.125
The different combinations of 1,3,6; 1,4,5; 2,2,6; 2,3,5; 2,4,4; 3,3,4 can produce the sum 10.For each of these there can be further arrangements within 3 dices.so 3P3 + 3P3 + 3P3/2! + 3P3 + 3P3/2! + 3P3/2! = 3*6 + 3*3 = 27
Total possible outcomes in the dice = 6^3 = 216
Therefore the required probability = 27/216 = 1/8 = 0.125

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ravikumar  (11 years ago)

three dices s0 6^3=216
from the three dices we are getting 27 numbers sum of 10
s0 ans will be
27/216

Is this solution Helpfull? Yes (1) | No   

Narendra Patel   (11 years ago)

if two dices are rolled then
p=3/36=1/12.
but if 3 dices are rolled then
n(s)=6*6*6;
n(a)={}
p(A)=n(a)/n(s)
pls solve anyone n(a) and put value in this

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Vivek Prajapati  (11 years ago)

1/18?

as 10 can be obtained with dice giving 1,5,4(6 ways) ad 2,5,3(6 ways)
so 12/216 i.e 1/18

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Guru Pranav  (11 years ago)

CASES :-
(6,3,1) - 6
(6,2,2) - 3
(5,3,2) - 6
(5,4,1) - 6
(4,4,2) - 3
(4,3,3) - 3
Total no of possibilites = 27
probability = 27/216.

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SURUCHI KUMARI  (11 years ago)

@ Debabrata Roy Chowdhury
ur process is right but u forgot to alter the last letter in ur ans, bcz u calculated EIRNGO as the 45th term not as 46th. so ans will be EIRNOG

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Raja  (10 years ago)

Possible ways :-
(6,3,1) - 6
(6,2,2) - 3
(5,3,2) - 6
(5,4,1) - 6
(4,4,2) - 3
(4,3,3) - 3
Total no of possibilites = 27
probability = 27/216 = 1/8;

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