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14.Probability
normal
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
$\frac{2}{{3}}$
Solution
$\mathrm{P}([(\mathrm{A} \cap \overline{\mathrm{B}}) \cup(\overline{\mathrm{A}} \cap \mathrm{B})] \cap \overline{\mathrm{C}})$
$=\frac{3}{4}-\left\{\frac{1}{3}+\frac{1}{6}\right\}=\frac{1}{4}$
Standard 11
Mathematics
