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STT 441
Quiz 2
Feb. 9, 2015
1. A closet contains 7 pairs of shoes. If 5 shoes are randomly selected, what
is the probability that there will be
(a) at least one complete pair?
[3pts]
We first compute the probability of no complete pairs. The samples
space
S contains all the combinations of 5 shoes. Thus thereare
14
7
outcome in S. Since there is no complete pair, there are
5
5
ways
to choose pairs and 2 ways to choose one shoe for each pair. So
7 5
2 outcomes contains no complete pair, which implies that
5
7 5
5 2
P ({at least one complete pair is 1}) = 1− 14 = 1−0.336 = 0.664.
5
(b) exactly one complete pair?
[3pts]
We first select pairs. There are 7 ways.
Then we select the rest of
6 3
shoes which are not paired. There are
2 ways. Hence,
3
7 × 63 23
= 0.560.
P ({exactly one complete pair}) =
14
5
2. Suppose that each of 5 men at a party throws his hat into the center of the
room. The hats are first mixed up, and then each man randomly selects
a hat. Let E ={none of the men selects his own hat} and Ei ={the ith
man selects his own hat} for i = 1, 2, . . . , 4, 5.
(a) Find the relationship between E and Ei s for i=1, 2, 3, 4 and 5;[4pts]
First note that E c ={at least one man select his own hat}. Thus
Ei ⊂ E c for all 1 ≤ i ≤ 5. Moreover, for any outcome ω ∈ E c , there
exists some i between 1 and 5 such that ω ∈ Ei . In words, this says
5
[
there must be someone who gets his own hat. Therefore, E c ⊂
Ei .
i=1
Combining these two, we have
Ec =
5
[
Ei
i=1
(b) Find P (Ei );
[3pts]
This is a matching problem. The sample space contains 5! sample
points. The event Ei contains 4! sample points. Then
P (Ei ) =
1
4!
=
5!
5
2
for i = 1, · · · , 5.
3. Five people, designated as A, B, C, D and E are arranged in linear order.
Assuming that each possible order is equally likely, what is the probability
that
(a) A and B are next to each other?
[2pts]
Consider A and B as a whole (say AB), there are 4! ways to order
AB, C, D and E, and 2! ways to order the A and B in the group.
Thus the probability for A and B are next to each other is
2
4!2!
= .
5!
5
(b) there are exactly two people between A and B?
[3pts]
3
choices for the two people between A and B, 2! orders
There are
2
for these two people, 2! ways to order A and B, and 2! ways to
order the A and B group and the remaining one person. Thus the
probability for exactly two people between A and B is
3
2
2 2!2!2!
=
.
5!
10
(c) C is between A and B?
[3pts]
5
There are
ways to choose the positions for A, B and C, and 2
3
way to order them such that C is between A and B. There are 2!
ways to order remaining D and E. Thus the probability such that C
is between A and B is
5
1
3 × 2 × 2!
= .
5!
3
3
4. A pair of fair dice is rolled. What is the probability of the following events?
(a) The sum of dice is larger than 10.
[3pts]
The sample space has 36 outcomes and the event A={the sum of dice
is larger than 10} contains 3 sample points. Then, we have
P (A) =
3
1
=
.
36
12
(b) At least one of the dice is greater than 4.
[3pts]
The sample space has 36 outcomes and the event A={at least one of
the dice is greater than 4} contains 20 sample points. Then, we have
P (A) =
20
.
36
(c) The second die lands on a higher value than does the first.
[3pts]
The sample space has 62 = 36 outcomes. For each outcome, it belongs to one of the following three events:
E0 = {the value of the second die equals the value of the first die},
E+ = {the value of the second die is higher than the value of the first die},
E− = {the value of the second die is lower than the value of the first die}.
By symmetry, we have P(E+ ) = P(E− ). Since P(E0 ) = 6/36, we
have
15
.
P(E+ ) =
36
4