Count qualifying chocolates, then qualifying boxes

Let X count chocolates, out of 20, that fail the minimum criterion. Under the source constant-probability independent-chocolate model, $X\sim B(20,0.02)$. Then $P(X\ge1)=1-P(X=0)=1-(0.98)^{20}=0.33239\ldots=0.332$ to 3 significant figures. The Teacher completes the count definition, both parameters and the complement method.

Let Y count boxes, out of 4, containing at least one failing chocolate. Use the independent-box model with success probability $q=1-(0.98)^{20}$ from (i), so $Y\sim B(4,q)$, written $B(4,0.33239)$ in the source. Now $P(Y=2)=\binom42q^2(1-q)^2=0.29545\ldots=0.295$ to 3 significant figures. This retains the Teacher count/parameter/probability completions. The source precision remark says to keep at least 5 significant figures in intermediate probabilities before a final 3-significant-figure answer; using the stored unrounded q is preferable.