Math Help - Venn Diagram of A,B and C

1. Venn Diagram of A,B and C

Can anyone tell me if this is the right Venn Diagram for denoting C the event that exactly one of A and B occurs?

Its the one on the right at this website:
Venn Diagram -- from Wolfram MathWorld

Also how would you write an expression for C in terms of unions, intersections and complement of A and B.

And then use that to find an expression of P(C) in terms of P(A),P(B) and P(A n B)

2. C is the event that exactly one of A and B occurs. You only need circles for A and B on the Venn Diagram

So it means that either:

A occurs (but not B)
or
B occurs (but not A)

So P(C) = P( $A$ n $B^c$) U P( $B$ n $A^c$)

3. Originally Posted by Janu42
C is the event that exactly one of A and B occurs. You only need circles for A and B on the Venn Diagram

So it means that either:

A occurs (but not B)
or
B occurs (but not A)

So P(C) = P( $A$ n $B^c$) U P( $B$ n $A^c$)
So P(C) = P( $A$ n $B^c$) U P( $B$ n $A^c$)[/QUOTE] would be the expression for C in terms of unions, intersections and complements of A and B.

How would I use this to derive an expression for P(C) in terms of P(A), P(B) and P(AnB).

Would it be like this?

P(C) = ( P(A) - P(A n B) ) u ( P(B) - P(A n B))

This is the question:

(a) draw a Venn diagram
(b) Write down an expression for C in terms of union, intersections and complement of A and B.
(c) Using result (b) derive an expression for P(C) in terms of P(A), P(B) and P(A nB). Give a detailed proof of the result.

I dont no what the difference for the expression of C and the expression of P(C) is?

4. Originally Posted by mathsandphysics
So P(C) = P( $A$ n $B^c$) U P( $B$ n $A^c$) would be the expression for C in terms of unions, intersections and complements of A and B.
How would I use this to derive an expression for P(C) in terms of P(A), P(B) and P(AnB).

Would it be like this?

P(C) = ( P(A) - P(A n B) ) u ( P(B) - P(A n B))

This is the question:
(c) Using result (b) derive an expression for P(C) in terms of P(A), P(B) and P(A nB). Give a detailed proof of the result.
$A~\&~B$ are events. So is $C=(A\cap B^c)\cup(B\cap A^c)= A\Delta B$ the symmetric difference of $A~\&~B$.
There the probability of $C$ depends of the probabilities of $A~\&~B$.

Thus $P(C)=P(A)+P(B)-2(A\cap B)$