In a certain population $10\%$ of the people are rich, $5\%$ are famous and $3\%$ are rich and famous. The probability that a person picked at random from the population is either famous or rich but not both, is equal to

A die is loaded in such a way that each odd number is twice as likely to occur as each even number. If $E$ is the event that a number greater than or equal to $4$ occurs on a single toss of the die then $P(E)$ is equal to

If $A$ and $B$ are two independent events, then $A$ and $\bar B$ are

Let $A$ and $B$ are two events and $P(A') = 0.3$, $P(B) = 0.4,\,P(A \cap B') = 0.5$, then $P(A \cup B')$ is

The probability that at least one of $A$ and $B$ occurs is $0.6$. If $A$ and $B$ occur simultaneously with probability $0.3$, then $P(A') + P(B') = $

If $\mathrm{A}$ and $\mathrm{B}$ are two events such that $\mathrm{P}(\mathrm{A})=\frac{1}{4}, \mathrm{P}(\mathrm{B})=\frac{1}{2}$ and $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{8}$ find $\mathrm{P}$ $($ not $\mathrm{A}$ and not $\mathrm{B})$