# Axioms Proving.

• Mar 14th 2009, 07:45 PM
panda*
Axioms Proving.
Let A, B and C be any three events. Show that

(i) $\displaystyle P(A) = P(B)$if and only if $\displaystyle P(A \cap B^{c}) = P(A^{c} \cap B)$
(ii) Given $\displaystyle P(A) = 0.5$ and $\displaystyle P(A \cup (B^{c} \cap C^{c})^{c})= 0.8$, determine $\displaystyle P(A^{c} \cap (B \cup C))$

Show that for any two events :
(i) If $\displaystyle A \subset B$, then $\displaystyle P(A^{c} \cap B) = P(B) - P(A)$
(ii) If $\displaystyle A \subset B$, then $\displaystyle P(A) \leq P(B)$
(iii) Why is it incorrect to assume that for some events A and B, $\displaystyle P(A) = 0.4, P(B) = 0.3$ and $\displaystyle P (A \cap B) = 0.35$?
• Mar 15th 2009, 12:41 AM
Moo
Hello,

I'll give you most of the solution, but you will have to fill in some steps ;)

Thing that may come in handy :
$\displaystyle P(M \cup N)=P(M)+P(N)-P(M \cap N) \quad (1)$
$\displaystyle \Rightarrow P(M \cap N)=P(M)+P(N)-P(M \cup N) \quad (2)$
$\displaystyle P(M^c)=1-P(M) \quad (3)$
$\displaystyle (M \cup N)^c=M^c \cap N^c \quad (4)$ (de Morgan's law)
$\displaystyle (M \cap N)^c=M^c \cup N^c \quad (5)$ (de Morgan's law)

Quote:

Originally Posted by panda*
Let A, B and C be any three events. Show that

(i) $\displaystyle P(A) = P(B)$if and only if $\displaystyle {\color{red}P(A \cap B^c)} ={\color{red}P(A^c \cap B)}$

\displaystyle \begin{aligned} P(A \cap B^c) &=P(A)+P(B^c)-P(A \cup B^c) \quad \text{ by (2)} \\ &=P(A)+1-P(B)-(1-P(A^c \cap B)) \quad \text{ by (3),(4)} \end{aligned}

$\displaystyle P(A \cap B^c)=P(A)-P(B)+P(A^c \cap B)$

$\displaystyle {\color{red}P(A \cap B^c)}-{\color{red}P(A^c \cap B)}=P(A)-P(B)$

From here, it should be very easy to conclude :)

Quote:

(ii) Given $\displaystyle P(A) = 0.5$ and $\displaystyle P(A \cup (B^{c} \cap C^{c})^{c})= 0.8$, determine $\displaystyle P(A^{c} \cap (B \cup C))$
I'm thinking on this one...

Quote:

Show that for any two events :
(i) If $\displaystyle A \subset B$, then $\displaystyle P(A^{c} \cap B) = P(B) - P(A)$
$\displaystyle B=A \cup (B \cap A^c)$
why ? because let's consider an element in B. It is contained in A, or it is contained in B, but not in A. This latter possibility gives the set $\displaystyle B \cap A^c$
You can also see that since A and $\displaystyle A^c$ are disjoint, then A and $\displaystyle B \cap A^c$ are disjoint.

Hence $\displaystyle P(B)=P(A \cup (B \cap A^c))=P(A)+P(B \cap A^c)$
And the conclusion follows.

Quote:

(ii) If $\displaystyle A \subset B$, then $\displaystyle P(A) \leq P(B)$
Use (i) :
$\displaystyle P(B)=P(A)+P(B \cap A^c)$
but since $\displaystyle P(B \cap A^c)$ is a probability, it's $\displaystyle \geq 0$

thus $\displaystyle P(B) \geq P(A)+0$

Quote:

(iii) Why is it incorrect to assume that for some events A and B, $\displaystyle P(A) = 0.4, P(B) = 0.3$ and $\displaystyle P (A \cap B) = 0.35$?
Because $\displaystyle A \cap B$ is included in $\displaystyle A$
By (ii), we should have $\displaystyle P(A \cap B) \leq P(A)$, which is not the case here ;)