smc

When tossed $n$ times, the probability of getting exactly 2 heads and the rest tails is $$\left(\genfrac{}{}{0ex}{}{n}{2}\right){\left(\frac{1}{4}\right)}^2{\left(\frac{3}{4}\right)}^{n-2}.$$ Similarly, the probability of getting exactly 3 heads is $$\left(\genfrac{}{}{0ex}{}{n}{3}\right){\left(\frac{1}{4}\right)}^3{\left(\frac{3}{4}\right)}^{n-3}.$$ Now set the two probabilities equal to each other and solve for $n$: $$\left(\genfrac{}{}{0ex}{}{n}{2}\right){\left(\frac{1}{4}\right)}^2{\left(\frac{3}{4}\right)}^{n-2}=\left(\genfrac{}{}{0ex}{}{n}{3}\right){\left(\frac{1}{4}\right)}^3{\left(\frac{3}{4}\right)}^{n-3}$$ $$\frac{n(n-1)}{2!}\cdot \frac{3}{4}=\frac{n(n-1)(n-2)}{3!}\cdot \frac{1}{4}$$ $$3=\frac{n-2}{3}$$ $$n-2=9$$ $$n=\boxed{11}$$