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December 24th, 2017, 12:09 PM   #1
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Post Which is true?

Assume there are three events, A, B, and C. Also, assume that P(A∨B)=0.5, and P(C)=0.5. Which of the following is true?


1. A and B are disjoint
2. A, B, and C are disjoint
3. P(C)>P(A∧B)
4. P(A∧B∧C)<0.5
5. None of these


I'm trying to understand and think of all possible probabilities to either prove or refute the options as correct or wrong. Help?
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December 24th, 2017, 01:35 PM   #2
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Originally Posted by backtobasics View Post
Assume there are three events, A, B, and C. Also, assume that P(A∨B)=0.5, and P(C)=0.5. Which of the following is true?


1. A and B are disjoint
2. A, B, and C are disjoint
3. P(C)>P(A∧B)
4. P(A∧B∧C)<0.5
5. None of these


I'm trying to understand and think of all possible probabilities to either prove or refute the options as correct or wrong. Help?
assuming that $P[A \cup B \cup C] = 1$

1) not enough information

2) not enough information, $A \cup B$ and $C$ are disjoint

3) no, if $A=B$ then $P[A \cap B] = 0.5 = P[C]$

4) Yes, $(A \cap B) \subseteq (A \cup B),~(A \cup B) \cap C = \emptyset$ so $(A \cap B \cap C = \emptyset) \Rightarrow P[A \cap B \cap C] = 0$
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December 24th, 2017, 09:19 PM   #3
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assuming that $P[A \cup B \cup C] = 1$

1) not enough information

2) not enough information, $A \cup B$ and $C$ are disjoint

3) no, if $A=B$ then $P[A \cap B] = 0.5 = P[C]$

4) Yes, $(A \cap B) \subseteq (A \cup B),~(A \cup B) \cap C = \emptyset$ so $(A \cap B \cap C = \emptyset) \Rightarrow P[A \cap B \cap C] = 0$

But aren't your statement 2 and 4 contradictory? You state not enough information to prove $A \cup B$ and $C$ are disjoint, while in statement 4 you state they are.

Also what to do in conditions where A , B and C are just some events in a bigger sample space where $P[A \cup B \cup C]$ != 1
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