CHAPTER 2 PROBABILITY EXERCISES ANSWER KEY

Exercise 2.1.1


1.  S = { 1,2,3,...,10}

Event of Interest = odd number is spun=A ={ 1,3,5,7,9} 

Complement of Event A = A^c={2,4,6,8,10}


2. S ={(1,1),(1,2)...,(6,6)}
   
   Event of Interest = both dice are thesame= A ={ (1,1),(2,2),(3,3),(4,4),(5,5),(6,6)} 

   Complement of Event A = A^c= {(1,2),(1,3),...,(5,6)}
   
3. S= { (no toppings),(onions),(extra cheese),(peppers),(onions,extra cheese), (onions,peppers),(extra cheese,peppers),
          (onions,extra cheese,peppers)}
        
   Event of Interest = pizza with two toppings = A ={ (onions,extra cheese), (onions,peppers),(extra cheese,peppers)}
   
   Complement of Event A = A^c={(no toppings),(onions),(extra cheese),(peppers),(onions,extra cheese,peppers)}


4. S ={ RRR,RRB,RBR,BRR,BBR,BRB,RBB,BBB}

   Event of Interest = Three red cards = A = {RRR}
   
   Complement of Event A = A^c ={ RRB,RBR,BRR,BBR,BRB,RBB,BBB}

Exercise 2.2.1

1.  A ={(1),(2)} 
    Some more probabilities defined in S

    a) P_1=0.01, P_2=P_3=0.25, P_4=0.1, P_5= 0.19, P_6= 0.2

       P(A) = 0.01 + 0.25 = 0.35
  
  
    b) P_1=0.19 ,P_2=0.5 , P_3=0.01, P_4=0.15, P_5= 0.05, P_6=0.10


       P(A) = 0.19 + 0.5 = 0.69
 
 
 **You can have some more different probability assignments,but make sure the sum of the probabilities
  sum up to one.
 
 
2) There are infinitely many; 

   For the sample space S = { 1,2,3,4,5,6}, we can assign different probabilities, P1,P2,P3...P6, for the outcomes,
   as long as the sum of these probabilities is 1.0.
   

3) Since we assume that the spinner is fair, the probability assignments for each outcome in 

    S = { 1,2,3,...,10}
    
    are equal. That is, P1=P2=...=P10 = (1/10)
    
   Event A = an odd number is spun = {1,3,5,7,9}
   
   P(A) = P1+ P3 +P5 +P7 +P9 = (5/10)
   

4) Please check on this problem.  P(any topping) = 1/2 ?


Exercise 2.3.1



1.   S = { (1,2),(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6),(4,5),(4,6),(5,6) }

     P(Sum of the two cards is 7) = P((1,6)) + P((2,5)) + P((3,4)) = 3/15
     
     
    The cards here were drawn without replacement,so outcomes like (1,1), (4,4)etc. are not possible
    ,whereas in the game of craps the outcomes (1,1),(4,4)--repeating outcomes are possible.The number of
    possible outcomes in the game of craps is 36.
      
   
  

Exercise 2.5.1




In the urn there are  30 blue balls(B), and 50 red balls(R)


1. P(getting two red balls AND one blue ball) = P(RRB) + P(RBR) + P(BRR) 
                          
                      =[(50/80)*(49/79)*(30/78)] + [(50/80)*(30/79)*(49/78)] + [(30/80)*(50/79)*(49/78)]
                      
                      = 0.4473
                     

 
2. P(getting the first two balls red) = P(RRR) + P(RRB)
 
                    = [ (50/80)*(49/79)*(48/78)] + [ (50/80)*(49/79)*(30/78)]
                    
                    = 0.3877
                    

3. Note : Draw 4 balls without replacement


   P( all balls are of the same color) = P(RRRR) + P(BBBB) 
   
                      = [(50/80)*(49/79)*(48/78)*(47/77)] + [(30/80)*(29/79)*(28/78)*(27/77)]
                      
                      = 0.1629
                      
                      
                      
4. P(three red cards) = P(RRR) = [(26/52)*(25/51)*(24/50)] = 0.1176


5. P(sum of the two cards is 7) = P(1,6) + P(2,5) + P(3,4) + P(4,3) + P(5,2) + P(6,1)

                                = [(1/6)*(1/5)] + [(1/6)*(1/5)] + [(1/6)*(1/5)] + [(1/6)*(1/5)] + [(1/6)*(1/5)] + [(1/6)*(1/5)]
                                
                            OR    = 6* [(1/6)*(1/5)] 
                            
                                = 1/5
                             
 
 
6. NOTE : This was asked in #1 already.

Exercise 2.6.1




1. Assuming a fair coin implies that P(H)=P(T) = 1/2 thus,

   P( HHHH) = (1/2)^4 = 1/16 = 0.0625
   

2. Assuming that the player's performance is constant--the same shooting accuracy each instance ,

    P(SSSSSS) = (0.7)^6 = 0.1176
 

3. The plane arrives safely if at least one of the engines are working

   Event that it will arrive safely given that one engine is not functioning = A ={ SSF,SFS,FSS}
   
   P(engine fails)=P(F) = 0.0001      P(engine is working)=P(S)= 0.9999
   
   P(A) = P(SSF)+ P(SFS) + P(FSS) = 3*[(0.9999)*(0.9999)*(0.0001)]= 0.0003
   

4.   If A and B are independent then it should be that
   
   P(A/B) = P(A) and P(B/A) =P(B)
   
   Now, if A and B cannot occur at the same time,

    P(A/B)=0, which is not equal to P(A)
    
    and 
    
    P(B/A)= 0, which is not equal to P(B) 
    
    (Note :unless A and B are null events?) 
 
    
   Thus, if A and B cannot occur at the same time, they are not independent.
     

5.  Not independent.The calls may be connected with each other (e.g. two calls may come from the same household)


6. Since the subjects were selected at random, the calls are surely independent.


7.Not independent.The responses may be related (e.g. the reporter interviews a group of related people in the mall)