In a quality check three inspectors examine the same batch independently. The probabilities that they notice a particular flaw are 0.6, 0.5 and 0.2. What is the probability that at least two of the three notice it?

The three probabilities differ, so the cases must be listed one by one rather than counted with a binomial formula. Exactly two notice the flaw in three ways: the first two but not the third, $0.6\times 0.5\times 0.8=0.24$; the first and third but not the second, $0.6\times 0.5\times 0.2=0.06$; the second and third but not the first, $0.4\times 0.5\times 0.2=0.04$. That is 0.34. All three notice it with probability $0.6\times 0.5\times 0.2=0.06$. Adding the disjoint cases gives 0.34 + 0.06 = 0.40. The option 0.34 stops at 'exactly two' and forgets that all three noticing also counts as at least two; 0.06 is 'all three' alone; and $0.84=1-0.4\times 0.5\times 0.8$ is the probability that at least ONE inspector notices the flaw.