If A and B are two events such that Pleft( {A cup B} right) = Pleft( {A cap B} right) , then i

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14.Probability

hard

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

$P\left( {A \cap B'} \right) = 0$

$P\left( {A' \cap B} \right) = 0$

$P\left( A \right) + P\left( B \right) = 1$

(JEE MAIN-2014)

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.

Standard 11

Mathematics

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(JEE MAIN-2024)

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normal

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