Compare an ordered third success with exactly three successes

Assume that whether one Type A component is faulty is independent of whether any other is faulty, and that each Type A component has the same fault probability .02.

Let $X\sim B(50,0.02)$. Then $q=P(X>1)=1-P(X\leq1)\approx0.2642286055$, printed as .26423 (5 s.f.) and .264 (3 s.f.). Use the unrounded q in later calculations.

The first four boxes must contain exactly two qualifying boxes and the fifth must qualify. With independent box outcomes, the probability is $q^2(1-q)^2q\,\dfrac{4!}{2!2!}=6q^3(1-q)^2\approx0.05992065432=0.0599$ (3 s.f.).

Let $Q\sim B(4,q)$ count qualifying boxes among the first four. Then the probability is $P(Q=2)q$, giving the same .0599. The source uses rounded q=.26423; its final result agrees.

For $Y\sim B(5,q)$, $P(Y=3)=10q^3(1-q)^2\approx0.09986775721=0.0999$ (3 s.f.).

The event in (iii) is a subset of the event in (iv): it imposes the additional requirement that the fifth box is the third qualifying one. Other orders also contribute positive probability to (iv), so its probability is greater. The two formulas have ratio 6/10=3/5, confirming the comparison.