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If $A$ and $B$ are two events such that $P\left( {A \cup B} \right) = P\left( {A \cap B} \right)$, then the incorrect statement amongst the following statements is
$A$ and $B$ are equally likely
$P\left( {A \cap B'} \right) = 0$
$P\left( {A' \cap B} \right) = 0$
$P\left( A \right) + P\left( B \right) = 1$
Solution
Let $A$ and $B$ be two events such that
$\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A} \cap \mathrm{B})$
and $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$
option (a) : since $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A} \cap \mathrm{B})$
(given)
therefore $A$ and $B$ are equally likely
Suppose option $(b)$ and option $(c)$ are correct.
$\therefore \mathrm{P}\left(\mathrm{A} \cap \mathrm{B}^{\prime}\right)=0$ and $\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}\right)=0$
$\Rightarrow \mathrm{P}(\mathrm{A})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})=0$
and $P(B)-P(A \cap B)=0$
$\Rightarrow P(A)=P(A \cap B)$
and $\mathrm{P}(\mathrm{B})=\mathrm{P}(\mathrm{A} \cap \mathrm{B})$
Thus $P(A)=P(B)=P(A \cap B)$
$=\mathrm{P}(\mathrm{A} \cup \mathrm{B})$
$[\because \text { Given } \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{P}(\mathrm{A} \cup \mathrm{B})]$
Also, we know
$\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$
$=\mathrm{P}(\mathrm{A} \cap \mathrm{B})+\mathrm{P}(\mathrm{A} \cap \mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$
$=\mathrm{P}(\mathrm{A} \cap \mathrm{B})$
which is true from given condition
Hence, option $(a),(b)$ and $(c)$ are correct.
