Lesson Elementary operations on sets help solving Probability problems

Elementary operations on sets help solving Probability problems

Problem 1

The set  (A' ∩ B)  is the set of those elements of B, that do not belong to A.


So  (A' ∩ B)  is simply the difference of two subsets  B  and  (A ∩ B) :


    (A' ∩ B) = B \ (A ∩ B),      (*)


where the symbol  " \ "  symbolizes subtraction of subsets.


From (*),  n(A' ∩ B) = n(B) - n(A ∩ B) = 60 - 10 = 50.    ANSWER

Problem 2

J, K, and L are events in sample space S. 
Pr(J)=0.35 
Pr(K)=0.18 
Pr(L)=0.24 
Pr(J ∩ K)=0.11 
Pr(J' ∩ L')=0.55      (*)
Pr(K' ∩ L)=0.16       (**)

What is Pr(J|K)? 

What is Pr(L|J)? 
 
What is Pr(K|L')?

    Pr(J|K) = by the definition = Pr(J ∩ K)/Pr(K) =  = .

    In the "given" part, find a line (*) related to L and J.

    This line is  Pr(J' ∩ L')=0.55

    From the elementary set theory,  J' ∩ L' = (S\J) ∩ (S\L) = S \ (J U L).   (Symbol " \ " means subtraction of sets.)

        You can prove it on your own using (or without using) Venn diagram for 2 subsets J and L in S.


    Also, from the basics of probability,  Pr(J U L) = Pr(J) + Pr(L) - Pr((J ∩ L).


                These pieces are THE KEY part of the solution, which you SHOULD and MUST understand when solving such problems !


    Therefore,  0.55 = Pr(J' ∩ L') = 1 - (Pr(J) + Pr(L)) + Pr((J ∩ L).  

    which implies  Pr(J ∩ L) = 0.55 + Pr(J) + Pr(L) - 1 = 0.55 + 0.35 + 0.24 - 1 = 0.14.


    Therefore, 

    Pr(L|J) = by the definition = Pr(L∩J)/Pr(K) =  = .

    By the definition,  Pr(K|L') = P(K ∩ L')/P(L'). 

    Pr(L') = 1 - Pr(L) = 1 - 0.24 = 0.76.

    What is  P(K ∩ L') ?


    In the "given" part, find a line (**) related to K and L.

    This line is  Pr(K' ∩ L)=0.16.


     The set   (K' ∩ L)  is  "L without K",  or,  in other words,  L\ (L ∩ K).

         Hence,  Pr((K' ∩ L) = 0.16 = Pr(L) - Pr(L ∩ K),   which gives

             Pr(L ∩ K) = Pr(L) - 0.16 = 0.024 - 0.16 = 0.08. 


     The set   (K ∩ L')  is  "K without L",  or,  in other words,  K\ (L ∩ K).

         Hence,  Pr(K ∩ L') = 0.18 - Pr(L ∩ K) = 0.18 - 0.08 = 0.10.


     Thus finally  Pr(K|L') =  = .

Problem 3

You may consider A and B as the subsets of the universal set U.



Then  (A ∩ B') = A \ (A ∩ B).       ( Elements of A that are not in B ) 

                                     The sign " \ " means subtraction of sets )


Therefore, you need first to find  P(A ∩ B)  and  then subtract it from P(A).



    Step 1.   P((A ∩ B) = P(A) + P(B) - P(AUB) =  +  -  =  = .


    Step 2.   P(A ∩ B') = P(A \ (A ∩ B)) = P(A) - P((A ∩ B) =  -  =  -  =  = .


Answer.  P(A ∩ B') = 

Problem 4

To calculate  P(A n B n C'), notice that  (A n B n C')  are those elements  (events)  of (A n B) that do not belong to C;

in other words,  (A n B n C') = (A n B) \ (A n B n C),      (1)

where the sign " \ " denotes subtraction of sets (of subsets).


Then equality (1) implies that

    P(A n B n C') = P(A n B) - P(A n B n C) = (substitute the given data) = 0.3 - 0.05 = 0.25.        ANSWER

Use the formula

    P(A U B U C) = P(A) + P(B) + P(C) - P(AnB) - P(AnC) - P(BnC) + P(A n B n C)     (2)

which is valid for any subsets  A, B and C  of the universal set U.


    This formula is very well known in elementary set theory.

    If you do not know its proof, here it is in few lines:

    
        Without my explanations, you understand well why I added P(A), P(B) and P(C) in the right side of the formula (2).

        But when I added these terms, I counted the intersections P(AnB), P(AnC)  and P(BnC) twice - so I need to subtract 
        these terms in the right side of the formula (2).

        But again,  when I added the terms P(A), P(B) and P(C) in the right side of the formula (2),
        I counted the intersection P(A n B n C) thrice. When I later subtracted the terms P(AnB), P(AnC)  and P(BnC), I fully compensate it,
        but now this intersection is not covered in the formula at all.
      
        Therefore I must add the term P(A n B n C), and this last step makes everything in equilibrium / (in balance).  

        The proof is completed.



    Now, when the formula (2) is proved, simply substitute the given input data in it. You will get

        P(A U B U C) = 0.5 + 0.6 + 0.4 - 0.3 - 0.1 - 0.2 + 0.05 = 0.95      ANSWER

To calculate  P(A' n B n C'), notice that  (A' n B n C')  are those elements  (events)  of B that do belong NEITHER  A  NOR  C;

in other words,  (A' n B n C') = (B \ (A n B) \ (C n B))    (3)



Then equality (2) implies that

    P(A' n B n C') = P(B \ (A n B) \ (C n B)) = (P(B) - P(AnB)) + (P(B) - P(CnB)) + P(A n B n C)     (4)


Notice that in the right side of the formula (4), I subtracted the probability P(AnB) from P(B), then subtracted P(CnB) from P(B), 
so I subtracted the probability P(AnC) of the intersection (AnC) twice; therefore, I should add P(A n B n C) to restore the equilibrium.


Substitute the given data into the formula (4), you get  the ANSWER

    P(A' n B n C') =   (0.6  - 0.3   - 0.2)   +   0.05       = 0.15.    (5)