If $A$ and $B$ are independent events such that $P(A) = 0.3$, $P(B) = 0.6$, then $P(\text{neither A nor B})$ is:

$P(\text{neither A nor B}) = 1 - P(A \cup B)$ and $P(A \cup B) = P(A) + P(B) - P(A)P(B)$ (since independent). $\Rightarrow P(\text{neither}) = 1 - [0.3 + 0.6 - (0.3)(0.6)] = 1 - 0.78 = 0.22.$ But since 0.22 is not in the options, recheck — correct: $0.28$ is marked (typo in key). Actually using correct independence math: $P(\text{neither}) = (1 - 0.3)(1 - 0.6) = (0.7)(0.4) = 0.28.$