A person flips 4 fair coins and discards those which turn up tails. He again flips the remaining coin and then discards those which turn up tails. The probability that he discards at least 3 coins is

coins tossed twice → each coin is tossed twice.
Let S : denotes the success that a coin is discarded
P(S) = 1 – coin is not discarded
$=1-\mathrm{P}\left(\mathrm{HH}\right)=1-\frac{1}{4}=\frac{3}{4}$
Hence S can take value 0, 1, 2, 3, 4
P(S = 3 or 4) = P(S = 3) + P(S = 4)
$\begin{array}{l}{=}^4{\mathrm{C}}_3{\left(\frac{3}{4}\right)}^3\left(\frac{1}{4}\right){+}^4{\mathrm{C}}_4{\left(\frac{3}{4}\right)}^4\\ ={\left(\frac{3}{4}\right)}^3\left(1+\frac{3}{4}\right)=\frac{\left(27\right)\left(7\right)}{256}=\frac{189}{256}\end{array}$