Suppose a random variable X follows the binomial distribution with parameters n and p, where `0 < p < 1` If `P(X=r)/(P(X=n-r))` is independent of n and r, then p equals

Explanation

Since `P(X=r) ^n C_r (p)^r (q)^(n-r) = n!/((n-r)!*r!) * (p)^r (1-p)^(n-r) ldots ldots [`Since `q=1-p] ldots (i)`

`P(X=0) = (1-P)^n`

And `P(X=n=r) = \ ^n C_(n-r) (P)^(n-r) (q)^(n-(n-r))`

`n!/((n-r)!r! * (p)^(n-r) ldots ldots [`Since` \ ^n C_r = \ ^n C_(n-r)] ldots (ii)`



`(P(X=r))/(P( x=n-r)) = [n!/((n-r)!r!) p^r (1-p)^(n-r)]/ [n!/((n-r)!r!) p^(n-r) (1-p)^(r)]`



`((1-p)/p)^(n-r) * 1/((1-p)/p)^r`