Let the man toss the coin n times. The n tosses are n Bernoulli trials Probability (p) of getting a head at the toss of a coin is $\frac{1}{2}$ ⇒ p = $\frac{1}{2}$ ⇒ q = $\frac{1}{2}$ ∴ P (X = x) = $\,{}^nC_x p^{n-x} q^x$ = $\,{}^nC_x \left(\frac{1}{2}\right)^{n-x} \left(\frac{1}{2}\right)^{x}$ = $\,{}^nC_x \left(\frac{1}{2}\right)^{n}$ It is given that P getting at least one head) > $\frac{80}{100}$ ⇒ P (X ≥ 1) > 0.8 ⇒ 1 - P (X = 0) > 0.8 ⇒ 1 - $\,{}^nC_n \frac{1}{2^n}$ > 0.8 ⇒ $\,{}^nC_0 \frac{1}{2^n}$ < 0.2 ⇒ $\frac{1}{2^n}$ < 0.2 ⇒ $2^n$ > $\frac{1}{0.2}$ = 5 ⇒ $2^n$ > 5 ... (1) The minimum value of n which satisfies the given inequality is 3. Thus, the man should toss the coin 3 or more than 3 times.