Q31 / 66 · JEE Advanced 2007 (Paper 2) · Mathematics

Let $E^c$ denotes the complement of an event $E$. Let $E, F, G$ be pairwise independent events with $P(G)>0$ and $P(E \cap F \cap G)=0$. Then, $P\left(E^c \cap F^c \mid G\right)$ equals

  • A
    $P\left(E^c\right)+P\left(F^c\right)$
  • B
    $P\left(E^c\right)-P\left(F^c\right)$
  • Correct answer
    C
    $P\left(E^c\right)-P(F)$
  • D
    $P(E)-P\left(F^c\right)$

Solution

$$ \text { } \begin{aligned} P\left(\frac{E^c \cap F^c}{G}\right) & =\frac{P\left(E^c \cap F^c \cap G\right)}{P(G)} \\ & =\frac{P(G)-P(E \cap G)-P(G \cap F)}{P(G)} \\ & =\frac{P(G)[1-P(E)-P(F)]}{P(G)} \\ & =1-P(E)-P(F) \\ & =P\left(E^c\right)-P(F) . \end{aligned} $$

Hence, the correct answer is option: 'C'.

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