Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that all the three persons apply for the same house is
- A$\frac{1}{9}$
- B$\frac{2}{9}$
- C$\frac{7}{9}$
- D$\frac{8}{9}$
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Let $A$ and $B$ be two events such that $P(A') = 0.3$,$P(B) = 0.4$,and $P(A \cap B') = 0.5$. Then $P(A \cup B')$ is:
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View SolutionLet $A$ and $B$ be two events such that the probability that exactly one of them occurs is $\frac{2}{5}$ and the probability that $A$ or $B$ occurs is $\frac{1}{2}$,then the probability of both of them occurring together is
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