Verified Solution • 2025–26 Session moderate

QJEEM16Madd83: JEE Main 2016 Mathematics - Class 12 JEEM16Madd83 Solution: If A and B are any two events such that P(A) = and P (A B) = ,…

If A and B are any two events such that P(A) = 25{2 \over 5} and P (A ∩\cap B) = 320{3 \over {20}} , hen the conditional probability, P(A ∣\left| {} \right. (A' ∪\cup B')), where A' denotes the complement of A, is equal to :

Multiple Choice Options

  • A 14{1 \over 4}
  • B 517{5 \over 17}
  • C 817{8 \over 17}
  • D 1120{11 \over 20}

Step-by-Step Verified Solution

P(AA′∪B′)P\left( {{A \over {A' \cup B'}}} \right)

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

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

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

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

As    A∩A′=ϕA \cap A' = \phi

∴\therefore     ϕ∪(A∩B′)=A∩B′\phi \cup \left( {A \cap B'} \right) = A \cap B'

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

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

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

Given   P(A)=25P\left( A \right) = {2 \over 5}

and    P(A∩B)=320P\left( {A \cap B} \right) = {3 \over {20}}

=25−3201−320= {{{2 \over 5} - {3 \over {20}}} \over {1 - {3 \over {20}}}}

=8−3201720= {{{{8 - 3} \over {20}}} \over {{{17} \over {20}}}}

=517= {5 \over {17}}

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