JEE Advanced · 2007 · 08 Apr 2007 · Paper 2 · Probability XII

Let

E^{c}

denote 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 (E^{c} \cap F^{c} ∣ G)

equals

Answer: (C)

Solution

Using the definition of conditional probability:

P (E^{c} \cap F^{c} ∣ G) = \frac{P ( E ^{c} \cap F ^{c} \cap G )}{P ( G )}

By De Morgan's law:

E^{c} \cap F^{c} = (E \cup F)^{c}

So:

P (E^{c} \cap F^{c} \cap G) = P (G) - P ((E \cup F) \cap G) = P (G) - P (E \cap G) - P (F \cap G) + P (E \cap F \cap G)

Since

P (E \cap F \cap G) = 0

and by pairwise independence

P (E \cap G) = P (E) P (G)

and

P (F \cap G) = P (F) P (G)

P (E^{c} \cap F^{c} \cap G) = P (G) - P (E) P (G) - P (F) P (G) = P (G) (1 - P (E) - P (F))

Therefore:

P (E^{c} \cap F^{c} ∣ G) = 1 - P (E) - P (F) = P (E^{c}) - P (F)

JEE Advanced · 2007 · 08 Apr 2007 · Paper 2 · Single correct option