Question

A coin is tossed $(m+n)$ times $(m<n)$ . The probability for getting atleast 'n' consecutive heads is

• A. $\frac{m+2}{2^{n+1}}$
• B. $\frac{n+2}{2^{m+1}}$
• C. $\frac{m}{2^{n+1}}$
• D. $\frac{(m+1)\times (m+2)}{2^{(m+n+1)}}$

Answer 1

The correct option is D $\frac{(m+1)\times (m+2)}{2^{(m+n+1)}}$

Probability of $n$ consecutive heads $=\frac{\frac{(m+1)!}{m!}}{2^{(m+n)}}=\frac{m+1}{2^{(m+n)}}$

Probability of $(n+1)$ consecutive heads $=\frac{\frac{\left(m\right)!}{(m-1)!}}{2^{(m+n)}}=\frac{m}{2^{(m+n)}}$
Continuing this way, we come to all heads.

Probability of getting all heads $=\frac{1}{2^{(m+n)}}$

Thus, the required probability

$=\frac{m+1}{2^{(m+n)}}+\frac{m}{2^{(m+n)}}+\frac{m-1}{2^{(m+n)}}+...+\frac{1}{2^{(m+n)}}=\frac{(m+1)\times (m+2)}{2^{(m+n)}\cdot 2}=\frac{(m+1)\times (m+2)}{2^{(m+n+1)}}$