Let minimum number of tosses be n.

 $\therefore$ Probability of atleast two heads = 1 – P (no head) – P (exactly one head)

$\Rightarrow \text{1 }–{\left(\frac{1}{2}\right)}^{\mathrm{n}}-{}^{\mathrm{n}}{\mathrm{C}}_1\mathrm{·}{\left(\frac{1}{2}\right)}^{\mathrm{n}-1}\mathrm{·}\left(\frac{1}{2}\right)\ge 0.\text{96}$

⇒ 1 – (n + 1) ${\left(\frac{1}{2}\right)}^{\mathrm{n}}\ge$ 0.96

⇒ (0.04) $\ge \frac{(\mathrm{n}+1)}{2^{\mathrm{n}}}$

⇒ 2ⁿ⁺² $\ge$ 100 (n + 1)

By hit and trial

$\therefore$ Minimum value of n will be 8.