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Solution :

consider the binomial distribution as`(1+p)^n` <br>
`P(x=m) = (.^nC_mP^m)/(1+p)^n` <br>
mean`= sum_m P(x=m)m= (sum .^nC_m.P^m*m)/(1+p)^n` <br>
`= P/(1+p)^n*sum.^nC_m . p_(m-1)m = n*p/(1+p) = 9` <br>
variance`= sum p(x=m)m^2 - (sum p(x=m)m)^2/9^2 ` <br>
`6 = (sum .^nC_mP^mm^2)/(1+ p)^n - 9^2` <br>
`6+9^2 = (sum.^nC_m P^m(m-1))/(1+p)^n + (sum m.^nC_m P^m)/(1+p)^n` <br>
` = p^2/(1+p)^n*n(n-1)(1+p)^(n-2) + (n*(1+p)^(n-1)p)/(1+p)^n` <br>
`= (n^2-n)*p^2/(1+p)^2 + np(1+p)/(1+p)^2` <br>
` 6+81 = (n^2p^2-np^2 +np +np^2)/(1+p)^2` <br>
`6 + 81 = (n^2p^2(n+np))/(np(1+p)^2)` eqn(1) <br>
`(np)/(1+p) = 9` <br>
`87= 81(1+np)/np` <br>
`87np = 81 + 81np` <br>
`26np = 81` <br>
`np = 27/2` <br>
`27/(2(1+p))= 9` <br>
`1+p = 3/2` <br>
`p=1/2` <br>
`n=27` <br>
`:. distribution (1+1/2)^27` <br>
`P(x=m)= (.^27C_m.(1/2)^m)/(3/2)^n` <br>
`= .^27C_m(1/3)^m(2/3)^(n-m)` <br>
`=(1/3 + 2/3)^m` <br>Related Video

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