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For any two events A and B in a sample space

  • Option 1)

    P\left ( \frac{A}{B} \right )\geq \frac{P(A)+P(B)-1}{P(B)}\cdot P(B)\neq 0,  does not hold.

     

     

     

     

     

  • Option 2)

    P(A\cap \bar{B})=P(A)-P(A\cap B) does not hold

  • Option 3)

    P(A\cup B)=1-P(\bar{A})P(\bar{B}), if A and B are independent

  • Option 4)

    P(A\cup B)=1-P(\bar{A})P(\bar{B}) , is A and B are disjoint

 

Answers (1)

best_answer

As we learned

 

Independent events -

P\left ( A \right )= P\left ( A\cap \overline{B} \right )\cup \left ( A\cap B \right )

P\left ( A\cap \overline{B} \right )= P\left ( A \right )\cdot P\left ( \overline{B} \right )

-

 

 

P(A\cup B)\leq 1\RightarrowP(A\cap B) ={P{A}+P(B)-P(A\cup B)}\geq {P{A}P(B)-1}

P\left ( \frac{A}{B} \right )=\frac{P(A\cap B)}{P(B)}\geq\frac{\left \{ P(A)+P(B)-1 \right \}}{P(B)} 

and 1-P(\bar{A})P(\bar{B})=1-(1-P(A))(1-P(B))=P(A)+P(B)-P(A)P(B)

=P(A)+P(B)-P(A\cap B)  (since A and B are independent)

=P(A\cup B)

a


Option 1)

P\left ( \frac{A}{B} \right )\geq \frac{P(A)+P(B)-1}{P(B)}\cdot P(B)\neq 0,  does not hold.

 

 

 

 

 

Option 2)

P(A\cap \bar{B})=P(A)-P(A\cap B) does not hold

Option 3)

P(A\cup B)=1-P(\bar{A})P(\bar{B}), if A and B are independent

Option 4)

P(A\cup B)=1-P(\bar{A})P(\bar{B}) , is A and B are disjoint

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Himanshu

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