Assume that $P(A)=0.4, P(A \cap B)=0.35$, and the probability that neither $A$ nor $B$ occurs is 0.3 . What is the probability of $B$ ?

Step 1 ：Let's denote the probability of event A as \(P(A)\), the probability of event B as \(P(B)\), the probability of both events A and B as \(P(A \cap B)\), and the probability of neither A nor B as \(P(\neg A \cap \neg B)\).

Step 2 ：We are given that \(P(A) = 0.4\), \(P(A \cap B) = 0.35\), and \(P(\neg A \cap \neg B) = 0.3\).

Step 3 ：We know that the probability of the union of two events A and B is given by the formula: \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\).

Step 4 ：We also know that the probability of the event that neither A nor B occurs is the complement of the event that either A or B occurs. Therefore, we have: \(P(A \cup B) = 1 - P(\neg A \cap \neg B)\).

Step 5 ：Substituting the given values into the equation, we get \(P(A \cup B) = 1 - 0.3 = 0.7\).

Step 6 ：Now, we can substitute the values of \(P(A)\), \(P(A \cap B)\), and \(P(A \cup B)\) into the formula for the union of two events to solve for \(P(B)\).

Step 7 ：So, \(0.7 = 0.4 + P(B) - 0.35\). Solving this equation for \(P(B)\), we get \(P(B) = 0.7 - 0.4 + 0.35 = 0.65\).

Step 8 ：Final Answer: The probability of B is approximately \(\boxed{0.65}\).