Multiple choice

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

$\displaystyle \frac{m+2}{2^{n+1}}$
$\displaystyle \frac{n+2}{2^{m+1}}$
$\displaystyle \frac{m}{2^{n+1}}$
$\dfrac { (m+1)\times(m+2) }{ { 2 }^{ (m+n+1) } } $

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