In two tosses of a coin, the sample space is given by
`S={HH,HT,TH,TT}.`
`therefore" "n(S)=4.`
So, every single outcome has a probability `(1)/(4).`
Let X= number of tails in two tosses.
In two tosses, we may have no tial. 1 tail or 2 tails.
So, the possible values of X are 0, 1, 2.
`P(X=0)=P("getting no tial")=P(HH)=(1)/(4).`
`P(X=1)=P("getting 1 tail")=P(HT or TH)=(2)/(4)=(1)/(2).`
`P(X=2)=P("getting 2 tails")=P(T T)=(1)/(4).`
Hence, the probability distribution of X is given by
`therefore" mean ",mu = Sigmax_(i)p_(i)=(0xx(1)/(4))+(1xx(1)/(2))+(2xx(1)/(4))=1.`
Variance, `sigma^(2)=Sigmax_(i)^(2)p_(i)-mu^(2)`
`=[(0xx(1)/(4))+(1xx(1)/(2))+(4xx(1)/(4))]-1^(2)`
`=(1)/(2).`
Standard deviation, `sigma=(1)/(sqrt(2))xx(sqrt(2))/(sqrt(2))=(sqrt(2))/(2)=(1.414)/(2)=0.707.`
