Please solve this question number is 146,147.Hurry up.

first if we throw three dice no of elements in samplespace will be 6*6*6= 216

we want all three different faces.

first get the same faces

same faces on all three dice (1,1,1) or (2,2,2) or (3,3,3) or (4,4,4) or (5,5,5,) or (6,6,6,) i.e. total 6 different ways.

now two same faces and one different face

starting with 1 i.e. two faces having one and third face other than one.

1,1,2 now these faces can be arranged in 3!/2! ways (there are three numbers out of which two are same so (3!/2!) = 3 ways

1,1,3 similarly here also 3 ways

1,1,4 similarly here also 3 ways

1,1,5 similarly here also 3 ways

1,1,6 similarly here also 3 ways

total for two face with 1 and third face with different num we have 3+3+3+3+3= 15 ways

similarly for two faces having 2 and one face having other than 2

(2,2,1) or (2,2,3) or (2,2,4) or (2,2,5) or (2,2,6) calculating in the same way as we have calculated for two faces 1 and one face with different num we can get 15 ways

similarly for two faces having 3 and one face having other than 3 = 15 ways

similarly for two faces having 4 and one face having other than 4 = 15 ways

similarly for two faces having 5 and one face having other than 5 = 15 ways

similarly for two faces having 6 and one face having other than 6 = 15 ways

so upon adding all these ways we get 90 ways .

so 90 ways of getting two similar faces and one different face

and 6 ways of getting all three with similar faces

total 96 ways of getting similar faces in any manner

so getting entirely different faces = 216-96= 120 ways

so prob is 120/216= 5/9