Question

A coin is tossed $(m+n)$ times, with $m>n$.The probability of getting $m$ consecutive heads is

• A. $\frac{(n+2)}{2^{m+1}}$
• B. $\frac{(m+2)}{2^{n+1}}$
• C. $\frac{(n+2)}{2^{m+n}}$
• D. None of these
  

Answer 1

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

$H=$ heads , $T=$ tail, $X=$ (head or fail)
Let $E=$ event of getting $M$ consecutive heads is the union of the following mutually exclusive events $A$; where $i$ $=0,1,2,...n$
$A_0:(HHH...M$ times $\left)T\right(XXX...(n-1)$ times$)$
$A_1:T(HHH...M$ times $\left)T\right(XXX...(n-2)$) times$)$
$A_2:XT(HHH...m$ times$\left)T\right(XXX...(n-3)$ times$)$
$A_{n-1}:(XXX...(n-2)$ times$\left)T\right(HHH...m$ times$)T$
$A_n:(XXX...(n-1)$ times$\left)T\right(HHH...m$ times$)$

It is easy to see that
$P\left(A_0\right)=P\left(A_n\right)={\left(\frac{1}{2}\right)}^m\frac{1}{2}$
and $P\left(A_1\right)=P\left(A_2\right)=...=P\left(A_{n-1}\right)$
$=\frac{1}{2}{\left(\frac{1}{2}\right)}^m\frac{1}{2}={\left(\frac{1}{2}\right)}^m=\frac{1}{4}$

Hence the reqiuerd probability is equal to
$P\left(E\right)=P\left(A_0\right)+P\left(A_1\right)+...P\left(A_{n-1}\right)+P\left(A_n\right)$

$\Rightarrow P\left(E\right)={\left(\frac{1}{2}\right)}^m\frac{1}{2}+{\left(\frac{1}{2}\right)}^m(\frac{1}{4}+\frac{1}{4}+...(n-1)$ $n$ times $)+{\left(\frac{1}{2}\right)}^m\left(\frac{1}{2}\right)$
$={\left(\frac{1}{2}\right)}^m(\frac{1}{2}+\frac{1}{2})+{\left(\frac{1}{2}\right)}^m\left(\frac{n-2}{4}\right)$
$={\left(\frac{1}{2}\right)}^m(\frac{1}{2}+\frac{1}{2})+{\left(\frac{1}{2}\right)}^m\left(\frac{n-2}{4}\right)$

$={\left(\frac{1}{2}\right)}^m(1+\frac{n-2}{4})=\frac{n+2}{2^{m+2}}$