Latest TCS Aptitude Question SOLUTION: Three dice are rolled. What is the probability that you will get the sum of the numbers as 10?

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(#M40024341) TCS Numerical Ability question Keep an EYE

Three dice are rolled. What is the probability that you will get the sum of the numbers as 10?

Asked In TCS MUKESH KUMAR CHAURASIA (10 years ago)

Unsolved Read Solution (5)

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Amaresh Viswanathan  (10 years ago)

n(s)=6^3=216.
no. of ways to get 10 as sum r
(2,2,6) -> 3!/2!=3.
(2,4,4) -> 3!/2!= 3
(1,3,6) -> 3!=6
(1,4,5) -> 3!=6
(2,3,5) -> 3!=6
(3,3,4) -> 3!/2!=3
Total n(e)=27.
Probability=n(e)/n(s)=27/216=1/8.

Is this solution Helpfull? Yes (28) | No (1)  

Harjit kaur  (10 years ago)

27/216
Let E be the event to get sum as 10
So E= {136,145,154,163,
226,235,244,253,262
316,325,334,343,352,361,
415,424,433,442,451
514,523,532,541
613,622,631}
N(E)= 27
now n(s) = 216. ( here s sample space )
so p(E)= 27/216=1/8

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amit kumar  (10 years ago)

total sample space=216
total event=27
so probability=27/216

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pala suphala  (10 years ago)

total sample space=216
total event=27
so probability=27/216

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Mahesh Gurung  (3 years ago)

basically its solving for equation x+y+z=10,
where x,y,z can take any value of 1,2,3,4,5,6;

solving this equation :

think x+y+z =10 like this,
number of ways we can distribute each 1 (10 one's ) by 2 sticks(sticks represent + sign)

e.g. 1 1 1 | 1 1 1 1 | 1 1 1
gap is filled by 2 sticks in this example making a possible permuation as 3+3+2;

1 Gap 1 Gap 1 Gap 1 Gap 1 Gap 1 Gap 1 Gap 1 Gap 1 Gap 1

so we can chose 2 gaps out of 9 gaps in 9C2 ways.

but
1+1+8
1+8+1
8+1+1
1+2+7 (permuate it in 3!=6 ways)
so 3 +9 of them must be subtracted out from 9C2.

9C2-9=27 ways.

probablity= 27/216.

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