Axioms Proving.

Hello,

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

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

Quote:

Originally Posted by panda*

Let A, B and C be any three events. Show that

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

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

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

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

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

Quote:

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

I'm thinking on this one...

Quote:

Show that for any two events :
(i) If $A \subset B$ , then $P(A^{c} \cap B) = P(B) - P(A)$

$B=A \cup (B \cap A^c)$
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 $B \cap A^c$
You can also see that since A and $A^c$ are disjoint, then A and $B \cap A^c$ are disjoint.

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

Quote:

(ii) If $A \subset B$ , then $P(A) \leq P(B)$

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

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

Quote:

(iii) Why is it incorrect to assume that for some events A and B, $P(A) = 0.4, P(B) = 0.3$ and $P (A \cap B) = 0.35$ ?

Because $A \cap B$ is included in $A$
By (ii), we should have $P(A \cap B) \leq P(A)$ , which is not the case here ;)