For a binomial variate X B(n, p) the difference between the mean and variance is 1 and the difference between their squares is 11 . If the probability of P(x=2)=m ( 5 6 )^n and n=36 then m: n=

Step-by-step solution

Given that difference of mean variance is 1 n p-n p q=1 n p(1-q)=1 n p^2=1 .... (i) Also given (n p)^2-(n p q)^2=11 .... (ii) n^2 p^2-n^2 p^2 q^2=11 n^2 p^2 (1-q^2 )=11 Divide (ii) by (i), we get aligned & n^2 p^2 (1-q^2 ) n p^2 =11 & n (1-q^2 )=11 1-q^2= 11 36 ( n=36) & q= 5 6 p= 1 6 aligned Now, p(x=2)=m ( 5 6 )^n aligned & ³⁶ C₂ ( 1 6 )^2 ( 5 6 )³⁴=m ( 5 6 )³⁶ & 36 35 2 ( 1 6 )^2 ( 5 6 )³⁴=m ( 5 6 )³⁶ & 36 7 10 5 6 5 6 ( 5 6 )³⁴=m ( 5 6 )³⁶ & = 36 7 10 ( 5 6 )³⁶=m ( 5 6 )³⁶ m= 36 7 10 aligned So, m: n= 36 7 10 :