If $A, B$ and $C$ are three events of a random experiment with $P(A)=0.4, P(B)=0.3$ and $P(A \cap B)=0.2$,then the probability that neither $A$ nor $B$ occurs is

$A$ lot consists of $12$ good pencils,$6$ with minor defects,and $2$ with major defects. $A$ pencil is chosen at random. The probability that this pencil is not defective is

$A$ natural number is selected at random from the set $\{x \in \mathbb{N} : 1 \leq x \leq 100\}$. The probability that the number satisfies the inequation $x^2 - 13x \leq 30$ is:

Let $A$ and $B$ be events for which $P(A) = x$,$P(B) = y$,and $P(A \cap B) = z$. Then $P(\bar{A} \cap B)$ equals:

Let $A$ and $B$ be events in a sample space $S$ such that $P(A)=0.5, P(B)=0.4$ and $P(A \cup B)=0.6$. Observe the following lists. The correct match of List $I$ from List $II$ is:

List $I$	List $II$
$(i) \ P(A \cap B)$	$(1) \ 0.4$
$(ii) \ P(A \cap \bar{B})$	$(2) \ 0.2$
$(iii) \ P(\bar{A} \cap B)$	$(3) \ 0.3$
$(iv) \ P(\bar{A} \cap \bar{B})$	$(4) \ 0.1$

Three persons work independently on a problem. If the respective probabilities that they will solve it are $\frac{1}{3}$,$\frac{1}{4}$,and $\frac{1}{5}$,then the probability that none of them can solve it is: