Results 1 to 2 of 2

Math Help - Axioms Proving.

  1. #1
    Junior Member
    Joined
    Nov 2008
    Posts
    45

    Axioms Proving.

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

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

    Show that for any two events :
    (i) If A \subset B, then P(A^{c} \cap B) = P(B) - P(A)
    (ii) If A \subset B, then P(A) \leq P(B)
    (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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    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* View Post
    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

    (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...

    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.

    (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


    (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
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: April 11th 2011, 11:26 AM
  2. Proving Theorems with Axioms
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: February 16th 2011, 06:36 PM
  3. [SOLVED] Field Axioms
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: September 6th 2010, 06:03 PM
  4. Axioms
    Posted in the Discrete Math Forum
    Replies: 5
    Last Post: September 10th 2009, 11:28 AM
  5. The Axioms of Arithmetic
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: February 21st 2007, 11:26 AM

Search Tags


/mathhelpforum @mathhelpforum