If A and B are any two events such that P(A)=2 /5 and P\left ( A\cap B \right )= 3/20, then the conditional probability,P\left ( A\mid \left ( A{}' \right \cup B{}') \right ),  where A{}' denotes the complement of A, is equal to :

  • Option 1)

    1/4

  • Option 2)

    5/17

  • Option 3)

    8/17

  • Option 4)

    11/20

 

Answers (1)
V Vakul

As we learnt in-

Demorgans Laws -

If A and B are any two sets then 

\Rightarrow \left ( A\cup B \right ){}'= A{}'\cap B{}'

\Rightarrow \left ( A\cap B \right ){}'= A{}'\cup B{}'

\Rightarrow P\left ( A_{1}\cap A_{2}\cap A_{3}\cdots \cap A_{n} \right )= P\left ( A_{1} \right )P\left ( \frac{A_{2}}{A_{1}} \right )P\left ( \frac{A_{3}}{A_{1}A_{2}} \right )\cdots P\left ( \frac{A_{n}}{A_{1}A_{2}A_{3}\cdots A_{n-1}} \right )

- wherein

where A1, A2.....An are n events

 

And, 

 

Conditional Probability -

 

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

and

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

 

- wherein

where P\left ( \frac{A}{B} \right ) probability of A when B already happened.

 

 P(A)=\frac{2}{5}\:.\:P(A\cap B)=\frac{3}{20}

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

Here, P({A}'\cup{B}')=P(A\cap B)' =1- \frac{3}{20}=\frac{17}{20}

P(A \cap (A \cap B)')=P(A)- P(A\cap B)

=\frac{2}{5}-\frac{3}{20}=\frac{5}{20}

P\left(\frac{A}{{A}'\cup{B}'} \right )=\frac{\frac{5}{20}}{\frac{17}{20}}=\frac{5}{17}

 


Option 1)

1/4

This option is incorrect.

Option 2)

5/17

This option is correct.

Option 3)

8/17

This option is incorrect.

Option 4)

11/20

This option is incorrect.

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