Question

A coin tossed $n$ times. If the probability that $4,5,6$ heads occur are in $A.P.$, then $n=$

• A. $14$
• B. $8$
• C. $15$
• D. $11$

Answer 1

Answer: (A)

$14$

$P(X=r)=\left(\begin{array}{c}n\\ r\end{array}\right){\frac{1}{2}}^r{\frac{1}{2}}^{n-r}=\left(\begin{array}{c}n\\ r\end{array}\right){\frac{1}{2}}^{n}P(X=4)=\left(\begin{array}{c}n\\ 4\end{array}\right){\frac{1}{2}}^4{\frac{1}{2}}^{n-4}P(X=5)=\left(\begin{array}{c}n\\ 5\end{array}\right){\frac{1}{2}}^5{\frac{1}{2}}^{n-5}P(X=6)=\left(\begin{array}{c}n\\ 6\end{array}\right){\frac{1}{2}}^6{\frac{1}{2}}^{n-6}P(X=5)=\frac{P(X=4)+P(X=6)}{2}\left(\frac{n}{5}\right)=\left[\right(\frac{n}{4})+(\frac{n}{6}\left)\right]/212\ast (n-5)=(n-5)\ast (n-4)+30n=7or14$