AIEEE 2007

JEE Main 2007 Mathematics question on Probability

Let EcE^c denotes the complement of an event EE. If EE, FF, GG are pairwise independent events with P(G)>0P(G) > 0 and P(E∩F∩G)=0P(E \cap F \cap G) = 0. Then, P(Ec∩Fc∣G)P(E^c \cap F^c | G) equals
  1. A.
    P(Ec)+P(Fc)P(E^c) + P(F^c)
  2. B.
    P(Ec)−P(Fc)P(E^c) - P(F^c)
  3. C.
    P(Ec)−P(F)P(E^c) - P(F)
  4. D.
    P(E)−P(Fc)P(E) - P(F^c)
Show answer and solution

Correct option: C

Solution

P(Eᶜ ∩ Fᶜ | G) = P(Eᶜ ∩ Fᶜ ∩ G) / P(G).

Now P(Eᶜ ∩ Fᶜ ∩ G) = P(G) - P(E ∩ G) - P(F ∩ G) + P(E ∩ F ∩ G).

Since E,F,G are pairwise independent: P(E ∩ G) = P(E)P(G) and P(F ∩ G) = P(F)P(G), and P(E ∩ F ∩ G) = 0.

So P(Eᶜ ∩ Fᶜ | G) = [P(G) - P(E)P(G) - P(F)P(G)] / P(G) = 1 - P(E) - P(F) = P(Eᶜ) - P(F), which is option 3.

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