GO Classes 2024 | Weekly Quiz 8 | Conditional Probability | Question: 1

Answer 1

$Red$ shaded area: $E \cap F^{c}$;

$Yellow$ shaded area: $E \cap F$;

$Orange$ shaded area: $F \cap E^{c}$;

$Green$ shaded area: $G$

Now, if $E$ & $G$ are $independent$ $events$, So, $P(E \cup G) = P(E) + P(G) – P(E \cap G)$ 

Where, $P(E \cap G) = P(E).P(G)$

If these two events are $mutually$ $exclusive$ then $P(E \cap G) = 0$

As two $mutually$ $exclusive$ events are $independent$ $iff$ $P(E) = 0$ or $P(G) = 0$

Then we can write this also, $P(E \cup G) = P(E) + P(G)$

Remember: $Independent$ $\neq$ $mutually$ $exclusive$

 $E, G$ are $mutually$ $exclusive$ = $P(E \cap G) = 0$

$E, G$ are $independent$ = $P(E \cap G) = P(E).P(G)$

$Ans: A;B$

Answer 2

• A. P(E U G) = P(E) + P(G) – P(E,G) [Inclusion Exclusion Principle]. In diagram we can see, E & G are mutually exclusive. So, P(E,G) = 0. Therefore, P(E U G) = P(E) + P(G).
• B. P(E U G) = P(E) + P(G) – P(E,G) [Inclusion Exclusion Principle].
• C. RHS :- P(F’) - P(E) = 1 - P(F) - P(E) = 1 - (P(F)+P(E)) = 1 - (P(E,F’)+2*P(E,F)+P(E’,F)) = 1 - P(E,F’) - 2*P(E,F) - P(E’,F) [(E,F’), (E,F), (E’,F) are all disjoint sets]. 
  LHS :- P(E’,F). LHS != RHS.
• D. RHS :- 1 – (P(F’) - P(E)) = 1 - P(F’) + P(E) = P(F) + P(E) = P(E,F’)+2*P(E,F)+P(E’,F)  [(E,F’), (E,F), (E’,F) are all disjoint sets]. 
  LHS :- P(E’,F). LHS != RHS.

Answer 3

Fo any two events E and G, P(E U G)=P(E)+P(G)-P(E∩ G).From the above Venn diagram (E∩G)=Φ So P(E∩ G)=0 From that P(E U G)=P(E)+P(G) Now P(E’∩F)=P(E U F) – P(E)=P(E)+P(F)-P(E∩F)-P(E)=P(F)-P(E∩F)≠ P(F’)-P(E) as P(F’)-P(E)=1-P(F)-P(E) and also P(E’∩F)≠ (1-(P(F’)-P(E))) as  (1-(P(F’)-P(E)))=1-P(F’)+P(E)=P(F)+P(E)

So Option A and B are correct .