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

2 votes

let Y be the event for getting total of 7

P(Y)=1/6

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)

=(5/6)^3

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

P(Y)=1/6

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)

=(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)

Case-3

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,

3*(1/6)(5/6)(5/6)+3*(1/6)(1/6)(5/6)+(1/6)(1/6)(1/6)

=75+15+1/216

= **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.**