For three non impossible events $A$, $B$ and $C$ $P\left( {A \cap B \cap C} \right) = 0,P\left( {A \cup B \cup C} \right) = \frac{3}{4},$ $P\left( {A \cap B} \right) = \frac{1}{3}$ and $P\left( C \right) = \frac{1}{6}$.
The probability, exactly one of $A$ or $B$ occurs but $C$ doesn't occur is
For three non impossible events $A$, $B$ and $C$ $P\left( {A \cap B \cap C} \right) = 0,P\left( {A \cup B \cup C} \right) = \frac{3}{4},$ $P\left( {A \cap B} \right) = \frac{1}{3}$ and $P\left( C \right) = \frac{1}{6}$.
The probability, exactly one of $A$ or $B$ occurs but $C$ doesn't occur is
- A
$\frac{1}{{12}}$
- B
$\frac{5}{{6}}$
- C
$\frac{1}{{4}}$
- D
$\frac{2}{{3}}$
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