A coin is tossed $$n$$ times. The probability of getting at least one head is greater than that of getting at least two tails by $$\frac{5}{{32}}.$$ Then $$n$$ is :

Solution :
The probability of getting at least one head
$$ = 1 - $$  probability of getting no heads
$$\eqalign{ & = 1 - {}^n{C_0}{\left( {\frac{1}{2}} \right)^0}.{\left( {\frac{1}{2}} \right)^n} \cr & = 1 - \frac{1}{{{2^n}}} \cr} $$
The probability of getting at least two tails
$$ = 1 - $$  probability of getting no tails $$-$$ probability of getting $$1$$ tail
$$\eqalign{ & = 1 - {}^n{C_n}{\left( {\frac{1}{2}} \right)^n}.{\left( {\frac{1}{2}} \right)^0} - {}^n{C_{n - 1}}{\left( {\frac{1}{2}} \right)^{n - 1}}.\frac{1}{2} \cr & = 1 - \frac{1}{{{2^n}}} - n\frac{1}{{{2^n}}} \cr} $$
From the question,
$$\eqalign{ & \left( {1 - \frac{1}{{{2^n}}}} \right) - \left( {1 - \frac{{1 + n}}{{{2^n}}}} \right) = \frac{5}{{12}} \cr & {\text{or }}\frac{{n + 1}}{{{2^n}}} - \frac{1}{{{2^n}}} = \frac{5}{{32}} \cr & {\text{or }}\frac{n}{{{2^n}}} = \frac{5}{{32}}\,\,\,\,\,\,\, \Rightarrow n = 5 \cr} $$