If $P(A) = \frac{1}{2}$,$P(B) = \frac{1}{3}$ and $P(A \cap B) = \frac{1}{4}$,then the value of $P(A' \cap B')$ is

Two integers $x$ and $y$ are chosen with replacement from the set $\{0, 1, 2, 3, \ldots, 10\}$. Then the probability that $|x-y| > 5$ is:

$A$ manager decides to distribute $Rs. 20000$ between two employees $X$ and $Y$. He knows $X$ deserves more than $Y$,but does not know how much more. So,he decides to arbitrarily break $Rs. 20000$ into two parts and gives $X$ the bigger part. Then,the chance that $X$ gets twice as much as $Y$ or more is

If $A$ and $B$ are events having probabilities $P(A)=0.6$, $P(B)=0.4$, and $P(A \cap B)=0$, then the probability that neither $A$ nor $B$ occurs is

The probability of choosing a vowel from the letters of the word $PROBABILITY$ is ........

For any event $A$,which of the following is true?