Verified Solution • 2025–26 Session moderate

QJEEAdv21Mae72b: JEE Advanced 2021 Mathematics - Class 12 JEEAdv21Mae72b Solution: Let E, F and G be three events having probabilities , and , and …

Let E, F and G be three events having probabilities P(E)=18P(E) = {1 \over 8} , P(F)=16P(F) = {1 \over 6} and P(G)=14P(G) = {1 \over 4} , and let P (E ∩\cap F ∩\cap G) = 110{1 \over {10}} . For any event H, if Hc denotes the complement, then which of the following statements is (are) TRUE?

Multiple Choice Options

  • A P(E∩F∩Gc)≤140P(E \cap F \cap {G^c}) \le {1 \over {40}}
  • B P(Ec∩F∩G)≤115P({E^c} \cap F \cap G) \le {1 \over {15}}
  • C P(E∪F∪G)≤1324P(E \cup F \cup G) \le {{13} \over {24}}
  • D P(Ec∪Fc∪Gc)≤512P({E^c} \cup {F^c} \cup {G^c}) \le {5 \over {12}}

Step-by-Step Verified Solution

For option (a)

P(E∩F∩GC)=P(E∩F)−P(E∩F∩G)≤P(E)−P(E∩F∩G)P(E \cap F \cap {G^C}) = P(E \cap F) - P(E \cap F \cap G) \le P(E) - P(E \cap F \cap G)

=18−110=140= {1 \over 8} - {1 \over {10}} = {1 \over {40}}

For option (b)

P(EC∩F∩G)=P(F∩G)−P(E∩F∩G)≤P(F)−P(E∩F∩G)P({E^C} \cap F \cap G) = P(F \cap G) - P(E \cap F \cap G) \le P(F) - P(E \cap F \cap G)

=16−110=115= {1 \over 6} - {1 \over {10}} = {1 \over {15}}

For option (c)

P(E∪F∪G)≤P(E)+P(F)+P(G)P(E \cup F \cup G) \le P(E) + P(F) + P(G)

=18+16+14=3+4+624=1324= {1 \over 8} + {1 \over 6} + {1 \over 4} = {{3 + 4 + 6} \over {24}} = {{13} \over {24}}

For option (d)

P(EC∩FC∩GC)=1−P(E∩F∩G)≥1−1324P({E^C} \cap {F^C} \cap {G^C}) = 1 - P(E \cap F \cap G) \ge 1 - {{13} \over {24}}

P({E^C} \cap {F^C} \cap {G^C}) \ge {{11} \over {24}} > {5 \over {12}}

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