Latest TCS Aptitude Question SOLUTION: Raj tossed three dice and their results are noted down. Find the total number of ways in which Raj can get the sum as 13. Also find the probability of get the sum as 10.

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Raj tossed three dice and their results are noted down. Find the total number of ways in which Raj can get the sum as 13. Also find the probability of get the sum as 10.

Asked In TCS subhrajit mitra (10 years ago)

Unsolved Read Solution (6)

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Read Solution (6) :

kratika jain  (10 years ago)

(6,6,1)=3!/2!=3
(6,5,2)=3!=6
(6,4,3)=3!=6
(5,5,3)=3!/2=3
(5,4,4)=3!/2!=3
total=21
P(sum 13)=21/216

Is this solution Helpfull? Yes (42) | No (2)  

tapas paul  (10 years ago)

Always remember when 3 dice are rolled the number of ways of getting n ( where n is the sum of faces on dice)
= (n−1)C2 where n = 3 to 8
= 25 where n = 9, 12
= 27 where n = 10, 11
= (20−n)C2 where n = 13 to 18
The required probability for sum 10 is = 27/6^3 = 27/216
For sum 13 (20-13)C2=7C2=21[ where n=13]
The required probability for sum 13 is = 21/6^3 = 21/216= 7/72

Is this solution Helpfull? Yes (33) | No (4)  

Ajay Poshak  (10 years ago)

number of ways in which 13 can appear on three dices
(6,6,1) = 3! / 2! = 3
(6,5,2) = 3! = 6
(6,4,3) = 3! = 6
(5,5,3) = 3! / 2! = 3
(5,4,4) = 3! / 2! = 3
(6,3,4) = 3! = 6
So the total number of ways in which 13 can appear will be 3+6+6+3+3+6 = 27

Is this solution Helpfull? Yes (7) | No (23)  

ira upadhyay  (10 years ago)

no of ways of 13=21;
p(10)=27/216;

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

payal sharma  (10 years ago)

i think it shud be 21*3/216

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sreehari  (9 years ago)

i think it is 18

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