(a) For \(X\sim\operatorname{Bin}(80,0.4)\), the mean is 32 and variance is 19.2. Since \(np=32\) and \(n(1-p)=48\) are large, use a normal approximation. Fewer than 27 means \(X\le26\), giving \[\operatorname{P}(X<27)\approx\Phi\!\left(\frac{26.5-32}{\sqrt{19.2}}\right)=0.105\quad\text{(3 s.f.)}.\]

(b) Let \(Y\sim\operatorname{Bin}(10,0.4)\). Then \[\operatorname{P}(Y\le2)=(0.6)^{10}+10(0.4)(0.6)^9+\binom{10}2(0.4)^2(0.6)^8=0.167\quad\text{(3 s.f.)}.\]