Obtain the probobility distribution of the number of sixes in two tosses of a fair die.

Explanation

When two dice are tossed Sample sapce, n (S) =36 Let X denote the no. of sixes
`:.X=0,1,2`
Case I : When X = 0 i.e, No sixes appear
`x={(1,1),(1,2),(1,3),(1,4),(1,5)`
`(2,1),(2,2),(2,3),(2,4),(2,5)`
`(3,1),(3,2),(3,3),(3,4),(3,5)`
`(4,1),(4,2),(4,3),(4,4),(4,5)`
`(5,1),(5,2),(5,3),(5,4),(5,5)}`
`:.n(X)=25`
`:.p(X=0)=(25)/(36)`
Case II : When X =1 i.e., only one six appears.
`x={(1,6),(2,6),(3,6),(4,6),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5)}`
`:.n(X)=10`
`:.=(X=1)=(n(x))/(n(s))=(10)/(36)`
Case III : When X =2 i.e., both coins show six.
`X={(6,6)}`
`:.n(X)=1`
`p(X=2)=(n(x))/(n(s))=(1)/(36)`
Thus probability distrbution of x is