Question

2K coins (K is an integer) each with probability P(O<P<1) of getting head are tossed together. If the probability of getting K heads is equal to the probability of getting K+1 heads, the value of P is

• A. $\frac{K}{2K+1}$
• B. $\frac{K+1}{2K}$
• C. $\frac{K+1}{2K+1}$
• D. $\frac{2K}{K+1}$

Answer 1

The correct option is C $\frac{K+1}{2K+1}$

Let $X$ be the number of heads occuring

$P(X=k+1)=\left(\begin{array}{c}2k\\ k+1\end{array}\right)p^{k+1}{(1-p)}^{k-1}$

Given

$P(X=k)=P(X=k+1)\left(\begin{array}{c}2k\\ k\end{array}\right)p^k{(1-p)}^k=\left(\begin{array}{c}2k\\ k+1\end{array}\right)p^{k+1}{(1-p)}^{k-1}\frac{k+1}{k}(1-p)=pp=\frac{k+1}{2k+1}$