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) = $\stackrel{\mathit{n}}{}{\mathit{C}}_{\mathit{x}}{\mathit{p}}^{\mathit{n}-\mathit{x}}{\mathit{q}}^{\mathit{x}}$ = $\stackrel{\mathit{n}}{}{\mathit{C}}_{\mathit{x}}{\left(\frac{1}{2}\right)}^{\mathit{n}-\mathit{x}}{\left(\frac{1}{2}\right)}^{\mathit{x}}$ = $\stackrel{\mathit{n}}{}{\mathit{C}}_{\mathit{x}}{\left(\frac{1}{2}\right)}^{\mathit{n}}$
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 - $\stackrel{\mathit{n}}{}{\mathit{C}}_{\mathit{n}}\frac{1}{2^{\mathit{n}}}$ > 0.8
⇒ $\stackrel{\mathit{n}}{}{\mathit{C}}_0\frac{1}{2^{\mathit{n}}}$ < 0.2
⇒ $\frac{1}{2^{\mathit{n}}}$ < 0.2
⇒ $2^{\mathit{n}}$ > $\frac{1}{0.2}$ = 5
⇒ $2^{\mathit{n}}$ > 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.