Probability — Class 12 Maths Solution

….(i)
$P(X = 0) = {(1 - p)^n}$

and $P(X = n - r){ = ^n}{C_{n - r}}{(p)^{n - r}}{(q)^{n - (n - r)}}$

….(ii)
Now, $\frac{{P(x = r)}}{{P(x = n - r)}} = \frac{{\frac{{n!}}{{(n - r)!r!}}{p^r}{{(1 - p)}^{n - r}}}}{{\frac{{n!}}{{(n - r)!r!}}{p^{n - r}}{{(1 - p)}^{ + r}}}}$

[using Eqs. (i) and (ii)]
$= {\left( {\frac{{1 - p}}{p}} \right)^{n - r}} \times \frac{1}{{{{\left( {\frac{{1 - p}}{p}} \right)}^r}}}$

Above expression is independent of $n$ and $r$, if $\frac{{1 - p}}{p} = 1 \Rightarrow \frac{1}{p} = 2 \Rightarrow p = \frac{1}{2}$