Question

Let \[E'\] denote the complement of an event \[E\]. Let \[E,F,G\] be pairwise independent events with \[P\left( G \right) > 0\], and \[P\left( {E \cap F \cap G} \right) = 0\]. Then \[P\left( {E' \cap F'|G} \right)\] equals
(A) \[P\left( {E'} \right) + P\left( {F'} \right)\]
(B) \[P\left( {E'} \right) - P\left( {F'} \right)\]
(C) \[P\left( {E'} \right) - P\left( F \right)\]
(D) \[P\left( E \right) - P\left( {F'} \right)\]

Answer 1

Hint: \[P\left( {E' \cap F'|G} \right)\] can be expanded using the formula of conditional probability. Thus, the result can be further simplified by taking \[P(G)\] common and canceling. \[P(E')\]is given by \[P(E') = 1 - P(E)\]. Using this, the result can be simplified further to obtain the required solution.

Formula Used:
If E and F are two events, then the conditional probability of E under the condition that F has occurred, written as \[P\left( {E{\text{ }}|{\text{ }}F} \right)\] is given by \[P\left( {E{\text{ }}|{\text{ }}F} \right) = \dfrac{{P(E \cap F)}}{{P(F)}},P(F) \ne 0\]. The probability that event \[E\] will not occur is denoted by \[P(E')\].\[P(E')\] is given by \[P(E') = 1 - P(E)\].

Complete step by step Solution:
\[P\left( {E' \cap F'|G} \right){\text{ }} = {\text{ }}\dfrac{{P\left( {E' \cap F' \cap G} \right)}}{{P\left( G \right)}}\]
                           \[ = {\text{ }}\dfrac{{P(G) - P(E \cap G) - P(G \cap F)}}{{P\left( G \right)}}\]
Since \[P(G) \ne 0\],
                           \[ = {\text{ }}\dfrac{{P(G)(1 - P(E) - P(F))}}{{P\left( G \right)}}\] [since E, F, G are pairwise independent]
                           \[ = (1 - P(E) - P(F))\]
                           \[ = {\text{ }}P\left( {E'} \right) - P\left( F \right)\]
Hence, \[P\left( {E' \cap F'|G} \right) = {\text{ }}P\left( {E'} \right) - P\left( F \right)\]

Hence, the correct option is (C).

Note: \[P\left( {E' \cap F'|G} \right){\text{ }} = {\text{ }}\dfrac{{P\left( {E' \cap F' \cap G} \right)}}{{P\left( G \right)}}\] only when \[P\left( G \right)\] is not equal to zero. If \[P\left( G \right) = 0\] then the value remains undefined.
Three or more events are said to be pairwise independent if they are independent when considered two at a time.