Find the probability distribution of the number of sixes in three tosses of a die.

Explanation

When a die is tossed we have S={1,2,3,4,5,6}
`:.` P(getting a six ) `=(1)/(6)` and P(not getting a six) `=(1-(1)/(6))=(5)/(6)`
Let X be the random variable denoting the number of sixes
In 3 trials the number of sixes may be 0 or 1 or 2 or 3
So, X may assume the values 0,1,2,3
`P(X =0)=.^(3)C_(0) ((1)/(6))^(0) .((5)/(6))^((3-0)) =(125)/(216)`
` P(X=1)=.^(3)C_(1) .((1)/(6))^(1).((5)/(6))^((3-1)) =(25)/(72)`
`P(X=2)=.^(3)C_(2).((1)/(6))^(2).((5)/(6))^((3-2)) =(5)/(72)`
`P(X=3)=.^(3)C_(3).((1)/(6))^(3).((5)/(6))^((3-3)) =(1)/(216)`
The required probability distribution X is given below:
`(underset(P(X): ,(125)/(216),(25)/(72),(5)/(72),(1)/(216))(X: ,0 , 1, 2, 3))`