Question 5
Mathematics
Probability
Easy
Let \( E^{c} \) denote the complement of an event \( E \) . Let \( E, F, G \) be pairwise independent events with \( P(G)>0 \) and \( P(E \cap F \cap G)=0 \) . The \( P\left(E^{c} \cap F^{c} \cap \mid G\right) \) equals
(A) \( P\left(E^{c}\right)+P\left(F^{c}\right) \)
(B) \( \mathrm{P}\left(\mathrm{E}^{\mathrm{c}}\right)-\mathrm{P}\left(\mathrm{F}^{\mathrm{c}}\right) \)
(C) \( \mathrm{P}\left(\mathrm{E}^{\mathrm{C}}\right)-\mathrm{P}(\mathrm{F}) \)
(D) \( P(E)-P\left(F^{c}\right) \)
View Dynamic Solution & Explanation ▼
Correct Solution: Option C
Step-by-step Solution:
We have \[ E \cap F \cap G = \emptyset \] \[ P(E^c \cap F^c \mid G) = \frac{P(E^c \cap F^c \cap G)}{P(G)} \] \[ = \frac{P(G) - P(E \cap G) - P(G \cap F)}{P(G)} \] From the Venn diagram: \[ E^c \cap F^c \cap G = G - E \cap G - F \cap G \] \[ = \frac{P(G) - P(E) P(G) - P(G) P(F)}{P(G)} \] (Since \(E, F, G\) are pairwise independent) \[ = 1 - P(E) - P(F) = P(E^c) - P(F) \]
