Download Discussion 12 Worksheet—Solutions

Discussion 12 Worksheet—Solutions
1.
a. Choose 2 of the 6 defects and 3 of the 9 non-defects. The answer is
 6  9 
  
 2  3 
15 
 
5 
.
b. At most 2 means exactly 2 or exactly 1 or exactly 0 defects. The answer is
 6  9   6  9   6  9 
          
 2  3  1  4   0  5 
15 
 
5 
.
c. The opposite of at least 2 defects is at most 1 defect (exactly 1 or exactly 0
defects). The answer is
 6  9   6  9 
      
1 4
0 5
1       
15 
 
5 
2.
1
1
 2 1
1
1
a. pE        
1  2   2 
2
2
2
 4 1
1
3
b. pF        
 2  2   2 
8
c. E  F  exactly one heads in first two tosses and exactly one heads in second
two tosses. Since the second two tosses are independent of the first two, the
probability of getting exactly one heads in the second two tosses is the same as
our answer to a). So we get pE  F     .
1 1
2 2
d. pE | F  
1
4
pE  F  1 4 2


pF 
38 3
e. pE   pF  
3 1
  pE  F  , so the events are not independent.
16 4
3.
a. There are 4 aces to choose from and we want 3 of them so
 4
 
3
p3 aces   
 52 
 
3
.
b. There are four aces, four kings and four queens and we want one of each, in
that order. So the desired probability is
444
.
P 52 ,3
Note the permutation in the
denominator reflects the fact that the hand is ordered.
c. The opposite of at least one ace being drawn is: no aces are drawn. Thus the
answer is
 48 
 
3
1  p no aces  1   
52
 
 
3
.
4. If heads is three times as likely as tails then for a single coin toss pheads  .75 and
10 
ptails   .25 . This is now a Bernoulli trials problem and the answer is  0.75 7 0.25 3 .
7
5. Define two events as follows: E  all four tosses result in heads and F  at least one
toss in four results in heads. Then we must find pE | F  . Now note that E  F hence
EF  E
pE | F  
and
so
pE  F   pE  
pE  F  1 16
1

 .
pF 
15 16 15
1
.
16
Also
pF   1  pF   1 
1 15
.

16 16
Thus
6. The expected value of the ping pong ball drawn is
1
2
1
1
3
27
.
1   2   3   4   5 
8
8
8
8
8
8
7. Because the questions are answered at random, each answer is independent of all the
others. The probability of getting a question correct is
question wrong. (Just like tossing a fair coin.)
7
3
10
10  1
10  1
1
a. pexactly 7 correct answers           .
 7  2   2 
 7  2 
1
, as is the probability of getting a
2
b. We need to have question #4 be one of the 7 correct answers plus six other correct
10 
answers chosen from the 9 other questions. There are   possible ways to get exactly 7
7

out 10 questions correct. So the desired probability is

9
  9
1     
 6   6 .
10  10 
   
7 7
c. Define the following events: E  question #4 is answered correctly, and F  exactly
seven questions are answered correctly. Then we must find pE | F  . Now E  F  #4 and
exactly six other questions are answered correctly. Since the individual questions are
1 9 1
answered randomly, we have pE  F      
2  6  2 
10
have pE | F  
 9  1 
  
 6  2 
10
10  1 
  
 7  2 
6
3
10
 9  1 
1
     
2
 
 6  2 
. Thus by part a) we
9
 
6
   .
10
 
 
7
8. Let’s call the containers A and B. In general, the probability of going free is
pgetting white | A chosen  pA chosen  pgetting white | B chosen  pB chosen .
Note that if the prisoner puts an equal number of white and black balls in A, then there
necessarily will be an equal number of each type in B. (For instance, 2 of each in A
implies 8 of each in B.) In any such case the probability of going free is
1 1 1 1 1
    .
2 2 2 2 2
Can the prisoner do better? As we just saw, if there is 1 of each ball in A and 9 of each
ball in B, the prisoner has equal odds of going free and not going free. What if we move
the one black ball from A to B. Then there is 1 white ball in A and the other 19 balls are
in B so the probability of going free is
1
1 9
 1    0.737 .
2
2 19
As it turns out, this is the best possible configuration. With this setup, it is of no use to
move any of the 9 white balls from B to A—this only lowers the probability of going
free. Likewise, moving black balls from B to A also lowers the chances of going free.