Download Homework 2 Solution

Homework 2 Solution
1. [§2-3] Two dices are thrown. Let E be the event that the sum of dice is
odd. let F be the event that at least one of the dice lands on 1, and let
G be the event that the sum is 5. Describle the events EF , E ∪ F , F G,
EF c and EF G
Solution:
EF ={the sum of the dice is odd and one of the dice lands on 1};
E ∪ F ={the sum of the dice is odd or at least one of the dice lands on 1};
EF c ={the sum of the dice is odd and no die lands on 1};
EF G={the sum of the dice is 5 and one of the dice lands on 1}.
2. [§2-6] A hospitcal administrator codes incoming patients suffering gunshot
wounds according to whether they have insurance (coding 1 if they do and
0 if they do not) and according to their condition, which is rated as good
(g), fair (f) or serious (s). Consider an experiment that consists of the
coding of such a patient
(a) Give the sample space of this experiment.
(b) Let A be the event that the patient is in serious condition. Specify
the outcomes in A.
(c) Let B be the event that the patient is uninsured. Specify the outcomes in B.
(d) Give all the outcomes in the event B c ∪ A.
Solution:
(a) The sample space is the collection of all the possible combinations of
insurance status and health condition.
S = {(1, g), (1, f ), (1, s), (0, g), (0, f ), (0, s)}.
(b) A = {(1, s), (0, s)}.
(c) B = {(0, g), (0, f ), (0, s)}.
(d) B c ∪ A = {(1, g), (1, f ), (1, s), (0, s)}.
3. [§2-9] A retail establishment accepts either the American Express or the
VISA credit card. A total of 24 percent of its customers carry an American
Express card, 61 percent carry a VISA card, and 11 percent carry both
cards. What percentage of its customers carry a credit card that the
establishment will accept?
1
Solution:
Let A ={customers carry an American Express card}, V ={customers
carry a VISA card}. Then P(A) = 0.24, P(V ) = 0.61 and P(A∩V ) = 0.11.
Thus
P({customers carry a credit card that the establishment will accept})
= P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 0.74.
4. [§2-12] An elementary school is offering 3 language classes: one in Spanish,
one in French, and one in German. The classes are open to any of the
100 students in the school. There are 28 students in the Spanish class, 26
in the French class, and 16 in the German class. There are 12 students
that are in both Spanish and French, 4 that are in both Spanish and
German, and 6 that are in both French and German. In addition, there
are 2 students taking all 3 classes.
(a) If a student is chosen randomly, what is the probability that he or
she is not in any of the language classes?
(b) If a student is chosen randomly, what is the probability that he or
she is taking exactly one language class?
(c) If 2 students are chosen randomly, what is the probability that at
least 1 is taking a language class?
Solution:
(a) Let S ={a randomly chosen student takes Spanish class}, F ={a
randomly chosen student takes French class}, and G ={a randomly
chosen student takes German class}. Then
P(S) = 0.28, P(F ) = 0.26, P(G) = 0.16,
P(S ∩ F ) = 0.12, P(S ∩ G) = 0.04, P(F ∩ G) = 0.06,
P(S ∩ F ∩ G) = 0.02.
Let E1 ={a randomly chosen student is not in any of the language
classes}. Then E1 = S c ∩ F c ∩ Gc .
P(E1 ) = P((S ∪ F ∪ G)c ) = 1 − P(S ∪ F ∪ G)
= 1 − [P(S) + P(F ) + P(G) − P(S ∩ F ) − P(S ∩ G) − P(G ∩ F )
+ P(S ∩ F ∩ G)]
= 0.5
2
(b) Let E2 ={a randomly chosen student takes at least two language
classes} and E3 ={a randomly chosen student takes exactly one language classes}. Then
P(E2 ) = P ((S ∩ F ) ∪ (S ∩ G) ∪ (G ∩ F ))
= P(S ∩ F ) + P(S ∩ G) + P(G ∩ F ) − 2P(S ∩ F ∩ G) = 0.18,
and
P(E3 ) = P(E1c ) − P(E2 ) = 0.32.
(There are many different ways to answer this question. We can
simply work it out by looking at the corresponding Venn diagram.)
(c) Let E4 ={at least 1 is taking a language class} and E4c ={neither one
is taking a language class}. From (a) we see that there 50 students
who do not take any language classes. Thus
50
49
2
=
,
P(E4c ) = 100
198
2
and
49
149
=
= 0.753.
198
198
52
5. [§2-15] If it is assumed that all
poker hands are equally likely, what
5
is the probability of being dealt
P(E4 ) = 1 −
(a) a flush? (A hand is said to be a flush if all 5 cards are of the same
suit.)
(b) one pair? (This occurs when the cards have denominations a, a, b,
c, d, where a, b, c, and d are all distinct.)
(c) two pairs? (This occurs when the cards have denominations a, a, b,
b, c, where a, b, and c are all distinct.)
(d) three of a kind? (This occurs when the cards have denominations a,
a, a, b, c, where a, b, and c are all distinct.)
(e) four of a kind? (This occurs when the cards have denominations a,
a, a, a, b.)
Solution:
(a) We
can divide the process into two stages: first
we choose a suit—
4
13
choices; then we choose five ranks—
choices. Therefore
1
5
the probability for a flush is
4 13
1
5
52
5
= 0.00198079231693.
3
13 4
(b) We first choose the pair—
choices. Then we select three
1
2
12 3
single cards—
4 . Therefore the probability for one pair is
3
13 4 12 3
4
1
2
3
= 0.42256902761104.
52
5
2
11 4
13
4
choices. The single card—
choices.
(c) Two pairs—
1
1
2
2
Therefore the probability for one pair is
42 4 11
13
2
2
1
1
52
5
= 0.04753901560624.
13 4
choices. Two single cards—
(d) Three cards of the same rank—
1
3
12 2
4 choices. Therefore the probability for three of a kind is
2
13 4 12 2
4
1
3
2
= 0.02112845138055.
52
5
12
13
choices. One single card—
4
(e) Four cards of the same rank—
1
1
choices. Therefore the probability for four of a kind is
13 12
4
1
1
= 0.00024009603842.
52
5
9. [§2-45] A woman has n keys, of which one will open her door.
(a) If she tries the keys at random, discarding those that do not work,
what is the probability that she will open the door on her kth try?
(b) What if she does not discard previously tried keys?
(Hint: be careful with the sample spaces in each case.)
Solution:
(a) If we use the sample space S of all permutations of n keys, then there
are n! outcomes in S. Let E ={open the door on the kth try}. Then
E contains permutations of n keys such that the right key is at the
kth position. Thus N (E) = (n − 1)! and
P(E) =
4
1
(n − 1)!
= .
n!
n
(b) Let the sample space S be the collection of all the possible outcomes
of m tries (m ≥ k). If we do not discard previously tried keys, the
total number of sample points in S is nm . And the number of the
sample points in the event E is (n − 1)k−1 nm−k . Thus
(n − 1)k−1 nm−k
=
P(E) =
nm
n−1
n
k−1 1
·
.
n
10. [§2-52] A closet contains 10 pairs of shoes. If 8 shoes are randomly selected,
what is the probability that there will be
(a) no complete pair?
(b) exactly 1 complete pair?
Solution:
(a) The samples
S contains all the combinations of 8 shoes. Thus
space
20
there are
outcomes in S. Since no shoes come from the same
8 10
ways to choose the pairs and 2 ways to choose
pair, there are
8
10 8
one shoe from each pair. Thus there are
2 different outcomes
8
such that there is no complete pair and the probability is
10 8
8 2
= 0.09145.
20
8
10
(b) There are
ways to choose one pair and 96 26 ways to choose 6
1
shoes such that there is no complete pair. Thus the probability for
exactly one pair is
10 9 6
1
6 2
= 0.4268.
20
8
5