If A and B are two events. The probability that at most one of A, B occurs, is

all the above

We have,

Required probability

$=P(\overline{A}∪\overline{B})= P(\overline{A ∪ B})= 1- P (A ∩ B)$

So, alternative (a) is correct.

Again

$P(\overline{A}∪ \overline{B})=P(\overline{A})+P(\overline{B})-P(\overline{A}∩ \overline{B})$ [By add. Theorem]

So, alternative (b) is correct.

Again,

$P(\overline{A}∪ \overline{B})$

$=P(\overline{A})+P(\overline{B})-P(\overline{A}∩ \overline{B})$

$=P(\overline{A})+P(\overline{B}) -P(\overline{A ∪ B})$

$=P(\overline{A})+P(\overline{B}) - \begin{Bmatrix}1- P(A ∪ B) \end{Bmatrix}$

$=P(\overline{A})+P(\overline{B}) + P(A ∪ B)-1$

So, alternative (c) is also correct.