Math/Stat 304
Solutions: TEST 1
Instructions: Do not simplify your answers --- particularly when they involve factorials,
combinations and/or permutations.
1. [5 pts] A fair four-sided die is thrown four times. Find the probability that the sum of the
numbers appearing is equal to 5.
Solution: Let the sample space  be the set of all 4-tuples from the alphabet {1, 2, 3, 4}. This
sample space is equiprobable and has cardinality 44.
Let E = “sum of four numbers is 5.” Now E consists of all 4-tuples having 3 ones and 1 two.
Thus #E = 4. So P(E) = 4/44 = 1/43.
2. [5 pts] A forest contains 20 elk, of which 5 are captured, tagged, and then released. One
month later, 4 of the 20 elk are captured. Find the probability that exactly 2 of these 4 have
been tagged. What assumptions have you made?
Solution: Let the sample space  be the set of all subsets of 4 of the 20 elk.
 20 
 .
4 
Then  is an equiprobable space of cardinality 
 5  15 
Let E = “2 of the four captured were tagged.” Then the cardinality of E is    .
 2 2 
 5 15 
  
# E  2  2 

Hence P(E) =
#
 20 
 
4 
2
3. [10 pts] In how many ways can 8 people be seated in a row if
(a) there are exactly 5 men and they must sit next to one another?
Solution: The 5 men can be “glued” together in 5! ways. This leaves 4 objects to
permute, which can be done in 4! ways. Thus the answer to the original question is
(4!)(5!).
(b) there are 4 married couples and each couple must sit together?
Solution: “Glue” each of the married couples together. This can be done in 24 ways.
Now permute the 4 couples; this can be done in 4! ways. Thus the answer to the original
question is 24(4!).
4. [5 pts] Albertine is dealt 26 cards from a well-shuffled standard deck of 52 cards. What is
the probability that she receives all four kings?
Solution: Let the sample space  be the set of all subsets of 26 cards from the deck of 52.
This is an equiprobable space. Let E = “4 kings included”. Once Albertine has the 4 kings
she must also have 22 of the remaining 48 cards.
 48 
.
 22 
Now #E = 
 48 
 
# E  22 

Hence P(E) =
#   52 
 
 26 
3
5. [10 pts] Vladimir has three vases (each of a different shape and color). Estragon presents
him with 25 flowers. In how many ways can Vladimir arrange the 25 flowers in the three
vases assuming that:
(a) each flower is unique (namely, different from the other 24 flowers)?
(Here you may assume that a vase (or even two) might remain empty.)
Solution: Each flower may be placed into any of the three distinguishable vases. The number
of ways this can be done is 325.
(b) the flowers are indistinguishable?
(Again, you may assume that a vase (or even two) might remain empty.)
Solution: Since the flowers are indistinguishable, we can simply count the number of
“code words” built from 25 stars and 2 slashes. Each code word corresponds to a unique
distribution of the 25 flowers into three distinguishable vases.
 27 
.
2 
So the number of such code words is 
6. [10 pts] A fair coin is flipped 25 times.
(a) Find the probability that exactly three heads appear.
Solution: Let the sample space  be the set of all sequence of length 25 from the
alphabet {H, T}.
This is an equiprobable sample space of cardinality 225.
 25 
.
3 
Let E be the event “exactly three heads.” Then # E = 
 25 
 
# E  3 
 25 .
Finally P( E ) 
#
2
4
(b) Find the conditional probability of exactly 3 heads appearing given that at least one
head appears.
Solution: Let F be the event “at least one head.”
Then P(F) = 1 – P(FC) = 1 
1
.
2 25
 25 
 
 25 
3 
 
25
3
P E  F  P ( E )
2
So P(E|F) = P( E | F ) 


 25 
P( F )
P( F ) 1  1
2 1
25
2
7. [5 pts] From a drawer that contains 10 pairs of gloves (each pair of a different style and
color), six gloves are randomly selected. Find the probability of obtaining exactly one pair of
gloves of the same style and color.
Solution: Let the equiprobable sample space  be defined as the set of all subsets of 6 gloves
from the given 20 gloves.
 20 
 .
6 
Then the cardinality of is 
Let E be the event “only one pair of gloves.” There are 10 ways of choosing the special pair.
Then one must select 4 gloves that don’t match. This is equivalent to first selecting 4 pairs of
the 9 that remain and then choosing the left or right glove of each selected pair.
9
10  2 4
4
9
#E
  
Then #E = 10  2 4 . Finally P( E ) 
#
 20 
 4
 
6 
5
8. [5 pts] Assume that each component has success probability of p.
Compute the probability of success for the given system.
Solution: The probability that the three components in parallel will fail is (1 – p)3. Thus this
subsysemt succeeds with probability 1 – (1 – p)3. Hence, because of independence, the
probability of the entire system succeeding is p3(1 – (1 – p)3).
9. [12 pts, MIT quiz problem] Archy and Mehitabel both wish to purchase a bicycle. The
bike store has a stock of four green, three yellow, and two red bikes. Archy randomly picks one
of the bikes and buys it. Immediately after, Mehitabel does the same. Let A be the event that
Archy bought a green bike, and B be the event that Mehitabel bought a green bike.
(a)
Find P(A).
Solution: We have P(A) = 4/9 (4 green bikes out of 9).
(b)
Find P(A|B).
Solution: P(A|B) = 3/8 (since we know that Mehitabel has a green bike, Archy can have one of
3 green bikes out of the remaining 8).
(c)
Are A and B independent events? Justify your answer.
Solution: Since P(A)  P(A|B), the events are not independent. Intuitively, since there is a fixed
quantity of green bikes, if Archy buys one, then the chances that Mehitabel buys one too are
slightly decreased.
Alternatively, P(A ∩ B) = P(A|B) · P(B) = (3/8)(4/9) = ½ but P(A)P(B) = (4/9)(4/9) = 16/81.
6
(d)
What is the probability that at least one of them bought a green bike?
Solution: The desired probability is P(A ∪ B). We know that
P(A ∪ B) = P(A) + P(B) − P(A ∩ B) =
P(A) + P(B) − P(A|B) · P(B) = 4/9 + 4/9 – (3/8)(4/9) ≈ 0.722
10. [10 pts] Seventy-five students, including Jules and Jim, are to be split into three classes of
equal size, and this is to be done at random. What is the probability that Jules and Jim end up in
the same class?
Solution: Let’s assume that the three classes are distinguishable: labeled A, B, C.
Then we can define an equiprobable sample space, , to be all distributions of the 75 students
into the three classes.

75

75!
 or
25! 25! 25!
 25 25 25 
The cardinality of  is then 
Let E denote the event “Jules and Jim are in the same class.”
Now Jules and Jim can be in any one of three classes. The remaining 73 students must be
distributed into classes of capacity 23, 25, 25. Thus the cardinality of E is
 73   50  25 
3     .
 23   25  25 
 73  50 
3   
23 25
Thus P( A)     . This simplifies to 12/37, just a bit smaller than 1/3.
 75 


 25 25 25 
7
11. [10 pts] A particular jury consists of 7 jurors. Each juror has a 0.15 chance of making
the wrong decision, independently of the others.
(a) If the jury reaches a decision by majority rule, what is that probability that it will reach the
wrong decision?
Solution: The chance of reaching the wrong decision can be achieved in any one of four
pairwise mutually exclusive ways: 4 wrong, 5 wrong, 6 wrong, or 7 wrong. This is equivalent
to flipping an unfair coin 7 times and asking for the probability of obtaining at least 4 tails
(where the probability of a tail is 0.15). Thus the answer is:
7
7
7
7
 (0.85)3 (0.15) 4   (0.85) 2 (0.15)5   (0.85)1 (0.15) 6   (0.85) 0 (0.15) 7
3 
 2
1 
0
(b) If two of the jurors have been bribed and threatened to give the “wrong” decision, what is
the probability that the jury will reach the wrong decision?
Solution: The easiest way to approach this problem, assuming independence, is to remove the
two “corrupted” jurors. Then 5 jurors remain. For a wrong decision to be made, the 5 jurors
must include 2 wrong, 3 wrong, 4 wrong or 5 wrong. These are pairwise mutually exclusive.
Thus the answer is:
5 
 5
5 
 5
 (0.85)3 (0.15) 2   (0.85) 2 (0.15)3   (0.85)1 (0.15) 4   (0.85) 0 (0.15)5
 2
 3
 4
 5
8
12. [15 pts] Odette has two stables on her farm: stable A houses 20 llamas and 13 moose;
stable B houses 4 llamas and 17 moose. Without any pattern, animals occasionally leave their
stables and then return to their stables.
(a) Assume that this morning, all the animals were in their assigned stables. What is the
probability that the next animal who wanders out is a llama?
Solution: Let SA = “animal leaves from Stable A” and SB = “animal leaves from Stable B”.
Let LA = “a llama leaves from Stable A” and LB = “a llama leaves from Stable B”.
Finally, let L = “a llama wanders out from a stable”
Then L = LA  S A   LB  SB  . Since the two sets in this union are mutually exclusive,
P( L)  PLA  S A   PLB  S B   PLA | S A P( S A )  PLB | S B P( S B ) 
(20 / 33)(1 / 2)  (4 / 21)(1 / 2)  0.3983
(b) Suppose that during a period when all the animals were in their stables, a llama comes
out of a stable and then returns to its dwelling. What is the probability that the llama
lives in stable A?
Solution: Here we use Bayes’ theorem. The prior probabilities for Stables A and B are
each ½. Now:
P ( S A | L) 
PS A  L  P( L | S A ) P( S A ) (20 / 33) (0.5)


 0.7608
P ( L)
P ( L)
0.3983
9
Extra Credit
Urn A contains 2 white and 4 red balls, whereas Urn B contains 1 white and 1 red ball. A
ball is randomly chosen from Urn A and put into Urn B, and a ball is then randomly
selected from Urn B. What is
(a) the probability that the selected ball from Urn B is white?
Solution: Let W2 = event that the ball chosen from Urn B is white.
Let W1 = event that the ball chosen from Urn A is white.
Note that P(W1) = 2/6.
Now W2  W2  W1   W2  W1C .
Note that the two sets in the union are disjoint. Hence




P(W2 )  PW2  W1   P W2  W1  PW2 | W1 P(W1 )  P W2 | W1 P(W1 ) 
C
C
C
(2 / 3)( 2 / 6)  (1 / 3)( 4 / 6)  4 / 9
(b) the conditional probability that the transferred ball was white given that a white ball
is selected from Urn B?
Solution: P(W1 | W2 ) 
PW1  W2  PW2 | W1 P(W1 ) 2 / 3(2 / 6)


 1/ 2
P(W2 )
P(W2 )
4/9