Question

Twenty coins each with probability $P(0<P<1)$ of showing head are tossed together. If the probability of getting $10$ heads is equal to the probability of getting $11$ heads, the value of $P$ is

• A. $\frac{9}{20}$
• B. $\frac{11}{20}$
• C. $\frac{11}{21}$
• D. $\frac{9}{21}$

Answer 1

The correct option is C $\frac{11}{21}$

$P(X=n)=\left(\begin{array}{c}20\\ n\end{array}\right)\ast p^n{(1-p)}^{20-n}\left(\begin{array}{c}20\\ 10\end{array}\right)\ast p^{10}{(1-p)}^{10}=\left(\begin{array}{c}20\\ 11\end{array}\right)\ast p^{11}{(1-p)}^9\frac{11(1-p)}{10}=pp=\frac{11}{21}$