JEE MAIN - Mathematics (2016 - 9th April Morning Slot - No. 14)

If A and B are any two events such that P(A) = $${2 \over 5}$$ and P (A $$ \cap $$ B) = $${3 \over {20}}$$, hen the conditional probability, P(A $$\left| {} \right.$$(A' $$ \cup $$ B')), where A' denotes the complement of A, is equal to :
$${1 \over 4}$$
$${5 \over 17}$$
$${8 \over 17}$$
$${11 \over 20}$$

Explanation

$$P\left( {{A \over {A' \cup B'}}} \right)$$

$$ = {{P\left[ {A \cap \left( {A' \cup B'} \right)} \right]} \over {P\left( {A' \cap B'} \right)}}$$

$$ = {{P\left[ {\left( {A \cap A'} \right) \cup \left( {A \cap B'} \right)} \right]} \over {P\left( {A \cap B} \right)'}}$$

$$\left[ \, \right.$$As   $$\left. {\left( {A \cap B} \right)' = A' \cap B'} \right]$$

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

As   $$A \cap A' = \phi $$

$$ \therefore $$   $$\phi \cup \left( {A \cap B'} \right) = A \cap B'$$

$$ = {{P\left( A \right) - P\left( {A \cap B} \right)} \over {1 - P\left( {A \cap B} \right)}}$$

$$\left[ \, \right.$$As   $$A \cap B' = A - \left( {A \cap B} \right)$$

and   $$\left. {\left( {A \cap B} \right)' = 1 - \left( {A \cap B} \right)} \right]$$

Given  $$P\left( A \right) = {2 \over 5}$$

and   $$P\left( {A \cap B} \right) = {3 \over {20}}$$

$$ = {{{2 \over 5} - {3 \over {20}}} \over {1 - {3 \over {20}}}}$$

$$ = {{{{8 - 3} \over {20}}} \over {{{17} \over {20}}}}$$

$$ = {5 \over {17}}$$

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