In a single tosses, we have
probability of getting a six `=(1)/(6),` and probability of getting a non-six `=(1-(1)/(6))=(5)/(6).`
Let X denote the number of sixes in two tosses.
Then, clearly X can assume the value 0, 1, or 2.
P(X=0)=P(non-six in the 1st draw) and (non-six in the 2nd draw)]
=P(non-six in the 1st draw) `xx`P(non-six in the 2nd draw)
`=((5)/(6)xx(5)/(6))=(25)/(36).`
P(X=1)=P[six in the 1st draw and non-six in the 2nd draw) or (non-six in the 1st draw and six in the 2nd draw)]
=P(six in the 1st draw and non-six in the 2nd draw)
+P (non-six in the 1st draw and six in the 2nd draw)
`=((1)/(6)xx(5)/(6))+((5)/(6)xx(1)/(6))=((5)/(36)+(5)/(36))=(10)/(36)=(5)/(18).`
P(X=2)=P[six in the 1st draw and six in the 2nd draw]
P=(six in the 1st draw) `xx` P (six in the 2nd draw)
`=((1)/(6)xx(1)/(6))=(1)/(36).`
Hence, the probability distribution is given by
`therefore" mean," mu=Sigmax_(i)p_(i)=(0xx(25)/(36))+(1xx(5)/(18))+(2xx(1)/(36))=(6)/(18)=(1)/(3).`
Variance, `sigma^(2)=Sigmax_(i)^(2)p_(i)-mu^(2)`
`=[(0xx(25)/(36))+(1xx(5)/(18))+(4xx(1)/(36))-(1)/(9)]=(5)/(18).`
Standard deviation, `sigma=sqrt((5)/(18))=(1)/(3)cdot sqrt((5)/(2)).`
