Suppose a random variable X follows the binomial distribution with parameters n and p, where `0

Options

A. `1/2`
B. `1/3`
C. `1/5`
D. `1/7`

Explanation

`becauseP(X=r)=""^(n)C_(r)(p)^(r )(q)^(n-r)=(n!)/((n-r)!r!)(p)^(r )(1-p)^(n-r)[becauseq=1-p]…(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)(1-p^(+r)[becauseq=1-p]because""^(n)C_(r)=""^(n)C_(n-r)]...(ii)`
Now, `(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))` [using eqs. (i) and (ii)]
`=((1-p)/p)^(n-r)xx1/((1-p)/p)^(r)`
Above expression is independent of n and r, if `(1-p)/p=1rArr1/p=2rArrp=1/2`