MHB Does the number of events in a set always sum to one?

[Dustinsfl]

If a single letter is selected at random from \(\{A, B, C\}\), find the probability of all events. Recall that the total number of events is \(2^n\), where \(n\) is the number of simple events. Do these probabilities sum to one? If not, why not?

This question comes verbatim from a book on probability and random process.

The probability of selecting a single letter is \(\frac{1}{3}\) and the probability of all events sums to \(1\). However, the question is then askig about power sets since it says recall that the total number of events is \(2^n\). This is the number of power sets. Aren't there only three events? Does it even make sense to sum power sets?

 

[zzephod]

If a single letter is selected at random from \(\{A, B, C\}\), find the probability of all events. Recall that the total number of events is \(2^n\), where \(n\) is the number of simple events. Do these probabilities sum to one? If not, why not?

This question comes verbatim from a book on probability and random process.

The probability of selecting a single letter is \(\frac{1}{3}\) and the probability of all events sums to \(1\). However, the question is then askig about power sets since it says recall that the total number of events is \(2^n\). This is the number of power sets. Aren't there only three events? Does it even make sense to sum power sets?

$A$ or $B$ is an event ,.,,

 

[Dustinsfl]

$A$ or $B$ is an event ,.,,

Both A and B are events as well as C.

 

[Jameson]

I can't really see the motivation of this question, especially with the restriction that only one letter may be chosen. If we can ignore that restriction for a moment, then it seems more pertinent as $P[A \cup B \cup C]=1$ and that calculation will involve all subsets of $\{A,B,C \}$. As you know though, this will not simply be a sum of all possible events because we need to subtract some events due to double counting. So if we add these together, this will be greater than 1.

How the above applies to choosing just one letter at a time though, that I can't say. Even so, here's my two cents.

 

[zzephod]

Both A and B are events as well as C.

I shall make things more explicit since a discreet prod is not sufficient: $\{A\}$, $\{B\}$ and $\{C\}$ are simple events and $(\{A\} \lor \{B\})$ is an event as is ...

.

 

[Dustinsfl]

I can't really see the motivation of this question, especially with the restriction that only one letter may be chosen. If we can ignore that restriction for a moment, then it seems more pertinent as $P[A \cup B \cup C]=1$ and that calculation will involve all subsets of $\{A,B,C \}$. As you know though, this will not simply be a sum of all possible events because we need to subtract some events due to double counting. So if we add these together, this will be greater than 1.

How the above applies to choosing just one letter at a time though, that I can't say. Even so, here's my two cents.

I shall make things more explicit since a discreet prod is not sufficient: $\{A\}$, $\{B\}$ and $\{C\}$ are simple events and $(\{A\} \lor \{B\})$ is an event as is ...

.

From these two post, I am inferring to use the Inclusion-Exclusion Principle; that is,
\[
P[A\cup B\cup C] = \sum P[E_i] - P[A\cap B] - P[B\cap C] - P[A\cap C] + P[A\cap B\cap C]
\]
where \(E_i\) is \(A\), \(B\), \(C\) for \(i = 1, 2, 3\)

 

[zzephod]

From these two post, I am inferring to use the Inclusion-Exclusion Principle; that is,
\[
P[A\cup B\cup C] = \sum P[E_i] - P[A\cap B] - P[B\cap C] - P[A\cap C] + P[A\cap B\cap C]
\]
where \(E_i\) is \(A\), \(B\), \(C\) for \(i = 1, 2, 3\)

All of the probabilities on the right hand side, other that the first sum, are zero (read your own question).

.

 

[Dustinsfl]

All of the probabilities on the right hand side, other that the first sum, are zero (read your own question).

.

I am sorry, but I have no idea what the hints are trying to steer me towards or what the question wants.

 

[Jameson]

I am sorry, but I have no idea what the hints are trying to steer me towards or what the question wants.

I really think is a strange question and would ask my professor to clarify. It's clear you understand the inclusion-exclusion principle and understand the power set, but how they tie together in this problem is strange.

Maybe the goal is to write out all possible events - no events, 1 event, 2 events and all three then notice that the probability of no events, 2 events and 3 events are all 0 so if you sum these you do in fact get 1.

 

[zzephod]

I am sorry, but I have no idea what the hints are trying to steer me towards or what the question wants.

It was not a hint but a statement, read your own question, can $A$ and $B$ both be drawn. The answer is no, so $P(\{A\} \cap \{B \})=0$ similarly any other intersection of distinct results has zero probability.

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