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2 votes

Probability of getting a total of 7 atleast once in three toss of a fair die is

  • 125/216
  • 91/216
  • 117/216
  • 9/216
in Probability
edited by
Is there any chance of getting sum '7' more than once..? Until we  tossed die 3 times ,we can't predict output and moreover die should be tossed 3times,so either sum will be '7' or not '7'. So what is the use of word at least once???
Atleast once means ??///
All of the ans given below are wrong, it should be 123/216

3 Answers

2 votes
let Y be the event for getting total of 7


Atleast once means 1-(none)

using binomial =nCx * p^x * q^(n-x)

take x=0 for none and n=3..since tossing 3 times

                        =3c0 * (1/6)^0 * (5/6)^(3)


so ans is 1-(5/6)^3 that is (91/216)
1 vote

The easiest way,

Probabilty of getting 7 is 1/6 [(1,6),(2,5),(3,4)]*2=6/36=1/6

p(Sum=7)=1/6 and thus, p(Sum is not 7)=5/6

Now, imagine 3 places,

_ _ _

Case-1: We get one time sum=7 ans two time not,

_7 _ _ 

so, there can be 3 cases, 3*(1/6)*(5/6)*(5/6)

Case-2: We get two time sum=7 ans one time not,

_7 _7  _ 

so, there can be 3 cases, 3*(1/6)*(1/6)*(5/6)


We get 7 at all positions,

_7 _7 _7 

and this is only one case, 1*(1/6)*(1/6)*(1/6)

Now, these all are 'OR' cases so  add them,



= 91/216

P.S. This is just a binomial method only, but in GATE during the process of solving questions remembering formulas can be hard and we solve all the questions iteratively like I did here.

0 votes
Trying brute fore method

when the die is tossed once ,probability of getting sum of 7 is 0

when the die is tossed 2 times,probability of getting sum of 7 is 6/36

when the die is tossed 3 times,probability of getting sum of 7 is 15/216

so total probability is (0+6/36+15/216)=51/216

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