If $A$ and $B$ are two events such that $P(A \cup B) = P(A \cap B)$,then the true relation is
- A$P(A) + P(B) = 0$
- B$P(A) + P(B) = P(A)P\left(\frac{B}{A}\right)$
- C$P(A) + P(B) = 2P(A)P\left(\frac{B}{A}\right)$
- DNone of these
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For two events $A$ and $B$,if $P(A) = P\left( \frac{A}{B} \right) = \frac{1}{4}$ and $P\left( \frac{B}{A} \right) = \frac{1}{2},$ then:
Medium
View SolutionIf $4\,P(A) = 6\,P(B) = 10\,P(A \cap B) = 1$,then $P\left( \frac{B}{A} \right) = \dots$
Easy
View Solution$A$ box contains $10$ mangoes, out of which $4$ are spoiled. $2$ mangoes are taken together at random. If one of them is found to be good, then the probability that the other is also good is:
Suppose $A$ and $B$ are events of a random experiment such that $P(A)=\frac{1}{3}$, $P(A \cap B)=\frac{1}{5}$ and $P(A \cup B)=\frac{3}{5}$. Match the items of List-$I$ with the items of List-$II$. List-$I$ List-$II$ $A$. $P(\frac{A}{B})$ $(i)$. $\frac{2}{15}$ $B$. $P(\bar{B})$ $(ii)$. $\frac{4}{15}$ $C$. $P(A \cap \bar{B})$ $(iii)$. $\frac{8}{15}$ $D$. $P(B \cap \bar{A})$ $(iv)$. $\frac{2}{3}$ $(v)$. $\frac{3}{7}$
| List-$I$ | List-$II$ |
| $A$. $P(\frac{A}{B})$ | $(i)$. $\frac{2}{15}$ |
| $B$. $P(\bar{B})$ | $(ii)$. $\frac{4}{15}$ |
| $C$. $P(A \cap \bar{B})$ | $(iii)$. $\frac{8}{15}$ |
| $D$. $P(B \cap \bar{A})$ | $(iv)$. $\frac{2}{3}$ |
| $(v)$. $\frac{3}{7}$ |
If $P(A / B) = \frac{3}{10}$, $P(B / A) = \frac{4}{5}$ and $P(A \cup B) = K P(B)$, then $\frac{1}{K} =$
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