Add a fixed box mass and retain both at-least-one methods

Let $A\sim N(1.52,0.070^2)$ be a pineapple mass in kg. Under the source independent-sampling model, $A_1+\cdots+A_6\sim N(1.52\times6,0.070^2\times6)=N(9.12,0.0294)$.

$P(A_1+\cdots+A_6+2.15<11.2)=P(A_1+\cdots+A_6<9.05)=0.341545\approx0.342$ to 3 significant figures. The fixed packaging adds no variance.

$P(\text{at least one below11.2kg})=1-P(\text{none below11.2kg})=1-(1-0.341545)^3\approx0.715$ to 3 significant figures. Distinct boxes are treated as independent with the same probability.

Let X count the filled boxes below 11.2 kg. Then $X\sim B(3,0.341545)$ and $P(X\ge1)=1-P(X=0)=0.714518\approx0.715$ to 3 significant figures. Both methods retain the source intermediate probability, rather than prematurely using 0.342.