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:

5

P(at least 1 head) = 1 – P (no head) = $1-{^nC}_0(\frac{1}{2})^n=1-(\frac{1}{2})^n$

P(at least 2 tails) = – P (no tails) – P(1 tail) = $1-{^nC}_n(\frac{1}{2})^n-{^nC}_{n-1}(\frac{1}{2})^{n-1}(\frac{1}{2})=1-(1+n)(\frac{1}{2})^n$

$[1-(\frac{1}{2})^n]-[1-(n+1)(\frac{1}{2})^n]=\frac{5}{32}⇒n(\frac{1}{2})^n=\frac{5}{2^5}⇒n=5$