[SOLVED] Probability Proof

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October 11th 2008, 08:38 AM
the1u2001

[SOLVED] Probability Proof

Hi i am new to this so can you please help with the following

Let A,B,C be events. Explain why

1. P(AuBuC)= P(A) + P(B/A) + P (C/(AUB)).

Deduce that

2. P(AUBUC)< P(A)+P(B) + P(C)

3. Formulate and prove a version of part (2) for n events.

Thank you
October 11th 2008, 10:19 AM
Plato

1 Attachment(s)

First, note that you mean $B\backslash A$ which is set difference.
What you typed looks like conditional probability.
The first one is just an exercise in sets.
Looking at the diagram, it is easy to see three disjoint sets. Their union is the whole.
$A \cup \left[ {B\backslash A} \right] \cup \left[ {C\backslash \left( {A \cup B} \right)} \right] = A \cup B \cup C$ .
So by the additive property $P(A) + P(B) + P(C) = P(A) + P\left[ {B\backslash A} \right] + P\left[ {C\backslash \left( {A \cup B} \right)} \right]$

For #2, use the monotone property of probability.
$T \subseteq S \Rightarrow \quad P(T) \leqslant P(S)$ (here again you typed < when it should be $\le$ ).
$\left[ {B\backslash A} \right] \subseteq B \Rightarrow \quad P\left[ {B\backslash A} \right] \leqslant P(B)$ .
A similar statement for the third term. Make the substitutions it follows.
October 11th 2008, 11:21 AM
the1u2001

Quote:

Originally Posted by Plato

First, note that you mean $B\backslash A$ which is set difference.
What you typed looks like conditional probability.
The first one is just an exercise in sets.
Looking at the diagram, it is easy to see three disjoint sets. Their union is the whole.
$A \cup \left[ {B\backslash A} \right] \cup \left[ {C\backslash \left( {A \cup B} \right)} \right] = A \cup B \cup C$ .
So by the additive property $P(A) + P(B) + P(C) = P(A) + P\left[ {B\backslash A} \right] + P\left[ {C\backslash \left( {A \cup B} \right)} \right]$

For #2, use the monotone property of probability.
$T \subseteq S \Rightarrow \quad P(T) \leqslant P(S)$ (here again you typed < when it should be $\le$ ).
$\left[ {B\backslash A} \right] \subseteq B \Rightarrow \quad P\left[ {B\backslash A} \right] \leqslant P(B)$ .
A similar statement for the third term. Make the substitutions it follows.

Thanks for the Help.

Sorry for typing the notations wrong but i didn't know how to post them (Giggle)
October 11th 2008, 11:34 AM
the1u2001

"For #2, use the monotone property of probability.
http://www.mathhelpforum.com/math-he...a7ff9f82-1.gif (here again you typed < when it should be http://www.mathhelpforum.com/math-he...2ece2e9c-1.gif).
http://www.mathhelpforum.com/math-he...7e6c79f7-1.gif.
A similar statement for the third term. Make the substitutions it follows"

Sorry to be a pain but i don't really understand what you mean by the monotone property of probability. Can you please exlain it to me.
October 11th 2008, 01:35 PM
Plato

If G is a subevent of H then the P(G) is less than or equal to P(H).
So $P(B\backslash A) \leqslant P(B)\,\& \,P\left( {C\backslash (A \cup B)} \right) \leqslant P(C)$ .
Thus $P(A \cup B \cup C) = P(A) + P(B\backslash A) + P\left( {C\backslash (A \cup B)} \right) \leqslant P(A) + P(B) + P(C)$ .