If $A$ and $B$ an two events such that $P\,(A \cup B) = \frac{5}{6}$,$P\,(A \cap B) = \frac{1}{3}$ and $P\,(\bar B) = \frac{1}{3},$ then $P\,(A) = $
If $A$ and $B$ an two events such that $P\,(A \cup B) = \frac{5}{6}$,$P\,(A \cap B) = \frac{1}{3}$ and $P\,(\bar B) = \frac{1}{3},$ then $P\,(A) = $
- A
$\frac{1}{4}$
- B
$\frac{1}{3}$
- C
$\frac{1}{2}$
- D
$\frac{2}{3}$
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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 $\mathrm{B})$
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Urn $A$ contains $6$ red and $4$ black balls and urn $B$ contains $4$ red and $6$ black balls. One ball is drawn at random from urn $A$ and placed in urn $B$. Then one ball is drawn at random from urn $B$ and placed in urn $A$. If one ball is now drawn at random from urn $A$, the probability that it is found to be red, is
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- [IIT 1988]
Suppose that $A, B, C$ are events such that $P\,(A) = P\,(B) = P\,(C) = \frac{1}{4},\,P\,(AB) = P\,(CB) = 0,\,P\,(AC) = \frac{1}{8},$ then $P\,(A + B) = $
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