114
Fall ECI 2011
Name
Solutions
Midterm 1 (100 points)
Wednesday, October 19, 2011
Show your work. Its not necessary to give a final numerical answer unless speci fically asked, but you need to
provide all the information necessary to compute the final answer. Please circle or otherwise highlight your
answer. Turn in your crib sheet with your exam. GOOD LUCK!
1. For each of the following statements, circle the letter T if it is true, and F if it is false.
(3 * 9 =27 pts) The correct answer has been bolded.
TF
a. S = { -2.0, -1.6, -1.2, -0.8, -0.4, 0.0, 0.4, 0.8, 1.2, 1.6, 2.0, } is an example
of a continuous sample space.
There are still only a countably infinite number of points in the sample space.
T
F
b. For any event A, (A S) (ACS) = S, where S represents the entire sample
space.
AS=A, ACS= AC, S=A AC by definition of complement.
T
F
c. Pr[A|Bc] = 1 Pr[A|B] .
For it to be true, the complement should be on the A: Pr[Ac|B] = 1 Pr[A|B].
TF
d.
If two events form a partition, then they are always independent.
Because they are M.E., and M.E. events are dependent knowing one occurs
affects the probability of the other occurring (you know it cannot occur, since
they are M.E.).
T F e. The mean and median are identical for symmetric distributions.
T
F
f. If Pr[A] + Pr[B] + Pr[C] = 1, then the set of events {A, B, C} forms a partition.
The converse of this is true. But suppose, e.g., B=C, and Pr[A] = Pr[B] =
Pr[C] = 1/3. Then the sum = 1, but {A, B, C} is not a partition since B and C
are not M.E. (nor are {A, B, C} C.E.). You could also devise an example in
which B and C overlap but are not equal.
n
n 1
T F g. 7 = n
7
6
LHS:
7 n!
n!
n (n 1)!
n!
=
=
, RHS:
7!(n 7)! 6!( n 7)!
6!(n 1 6)! 6!( n 7)!
1
T F h For a certain sample of 15 observations and a mean of 10, it is known that
i yi2 = 1584. The variance of this sample is 5.6.
yi 2 N y 2 1584 15(10) 2 84
S=
=
=
=6
N 1
14
14
2
TF
i. If A and B form a partition and B and C are collectively exhaustive, then
AC=A.
A B=S and A B= ; B C=S. Its easy to find C A , so A C=A.
A
B
C
Put a different way: If A C A, then there must be a point in A that is not in C. But if its in
A, then its not in B, since A and B are M.E. (since they are a partition). If its not in C and not in
B, then B and C cant be C.E., which is a contradiction.
2. At the coffee house in Memorial Union, people who bought hot drinks were observed last
Friday. Assume they bought either coffee or tea, but not both. It was found that 65% of all
customers bought coffee. 30% of the coffee drinkers put sugar in their drinks, whereas 60%
of the tea drinkers put sugar in theirs. A customer is selected at random. (3 + 5 + 5 + 5 + 5 +
5 +5 + 5 = 38 pts)
a. Define all relevant events, and express all numbers given in the problem statement as
probabilities of events.
C: The customer bought coffee
Pr[C]=0.65
T: The customer bought tea
Pr[T]=1-0.65=0.35
S: The customer put sugar in the beverage
Pr[S|C]=0.3, Pr[S|T]=0.6
C
(S : The customer did not put sugar in)
b. What is the probability that the customer put sugar in the beverage?
2
We know events C and T are a partition. Using the total probability formula, we get:
Pr[ S ] = Pr[ S | C ] Pr[C ] + Pr[ S | T ] Pr[T ] = 0.3 * 0.65 + 0.6 * (1 0.65) = 0.405
c. If the customer used sugar, what is the probability that s/he bought tea?
Using Bayes Theorem,
Pr[T | S ] =
Pr[ S | T ] Pr[T ]
Pr[ S | T ] Pr[T ]
0.6 * (1 0.65)
=
=
= 0.519
Pr[ S ]
Pr[ S | C ] Pr[C ] + Pr[ S | T ] Pr[T ]
0.405
d. Are the events buying coffee and putting sugar in the beverage dependent?
Explain why or why not, with words and with an equation.
Knowing someone is buying coffee may decrease the probability of putting sugar in the
beverage because the probability of a coffee drinker putting sugar in it (30%) is much
smaller than that of a tea drinker (60%). So our intuition is that the two events are
dependent.
To conduct a mathematical test for independence, we need to compare Pr[S|C] and
Pr[S]. From parts a) and b), we can easily see Pr[S|C]=0.3 that < Pr[S]=0.405. Therefore
the two events are dependent. (We could also note that 0.519 = Pr[T|S] < Pr[T] = 0.35).
It is also observed that 25% of all the customers put milk in their beverage. If the
customer bought coffee and put sugar in it, the probability of putting milk in as well is
90%.
e. What is the probability that the customer bought coffee and put both sugar and milk
in it?
First define the event:
M: putting milk in the beverage
Using the Chain Rule:
Pr[M]=0.25 Pr[M|C, S]=0.9
Pr[C S M ] = Pr[C ] Pr[ S | C ] Pr[ M | C , S ] = 0.65 * 0.3 * 0.9 = 0.1755
f. What is the probability that the customer bought coffee and put sugar in it, given that
she put milk?
From the conditional probability formula, we get:
Pr[C S ] = Pr[C ] Pr[ S | C ] = 0.65 * 0.3 = 0.195
Then use Bayes Theorem again:
3
Pr[C S | M ] =
Pr[C S ] Pr[ M | C S ] 0.195 * 0.9
=
= 0.702
Pr[ M ]
0.25
Drawing base Venn diagrams that reflect all relevant events (for the whole problem, not
just the events mentioned below) in their proper relation to each other, show on the base
Venn diagram and express in set notation the events that:
g. A customer bought coffee with milk, but not sugar in it.
C M S C
The shaded area.
C
T
S
M
h. A customer bought neither coffee with milk, nor tea with sugar.
(C M ) C (T S ) C = ((C M ) (T S )) C = (C MC) (T SC).
The total area except the shaded area.
C
T
S
M
4
3. 5 players are selected from a 20-student class to play a game. We know that 8 students in the
class are sophomores. (3+3+4+5 = 15 pts)
a. How many different ways are there to select 5 players from the class, assuming the 5 players
have identical roles in the game? (Final answer required).
Use the combinations rule:
20
( 520 ) = 15!*!5! = 15,504
b. How many ways in part (a) contain exactly 4 sophomores?
Use the multiple combinations rule:
( )( ) = 70 *12 = 840
8
4
12
1
c. What is the probability that at least 3 players selected are sophomores?
Use the multiple combinations rule again:
( )( ) + ( )( ) + ( )( ) = 56 * 66 + 70 *12 + 56 *1 = 0.296
15504
()
8
3
12
2
8
4
12
1
20
5
8
5
12
0
d. If we know among the final selected 5 players that 2 are freshmen, 2 are sophomores, and 1
is a junior, how many different ways can the 5 players be arranged in a row if only their class
level will be distinguished?
Use the partitions rule:
5!
= 30 .
2!*2!*1!
Alternatively, you can think of it as a series of combinations (because only their class level
will be distinguished, so we use combination instead of permutation): there are
( ) ways to
5
3
assign the 3 sophomores; then there are 3 positions left for the 2 freshmen, so there are
ways to assign them; at last, there is 1 position left, so
()
3
2
( ) way for assigning the junior
1
1
student. Therefore, the total number of ways is:
( ) * ( ) * ( ) =10*3*1=30.
5
3
3
2
1
1
5
4. During a recent sale at a music store, the number of CDs customers purchased was
distributed as follows. (8+2+5+5=20 pts)
Number of CDs X
Probability
0
1
2
3
a.
0.1
0.3
0.4
p
What is the expected number of CDs purchased, and its variance (first, obtain the value of
p)? (Final numerical answers required). (8 pts)
p( x ) = 1 , so p = 1 - 0.1 - 0.3 - 0.4 = 0.2.
i
i
4
E(X) = xi * p ( xi ) = 0 * 0.1 + 1 * 0.3 + 2 * 0.4 + 3 * 0.2 = 1.7 .
i =1
Var(X) = E[X2]-(E[X])2 =
4
x
i =1
2
i
* p ( xi ) 1.7 2 = (0 2 * 0.1 + 12 * 0.3 + 2 2 * 0.4 + 3 2 * 0.2) 1.7 2 = 0.81 .
b. What is the probability that a randomly-selected customer purchased fewer than 2 CDs? (2
pts)
p(xi=0)+p(xi=1) = 0.1+0.3 = 0.4.
c.
Completely specify the CDF for X. (5 pts)
CDF for X
Number of CDs (x)
x<0
0 x <1
1 x <2
2 x <3
x3
F(x)
0.0
0.1
0.4
0.8
1.0
d. Let I = 12 X - 6 (revenue minus cost) be the total net income of the store, in dollars, from a
randomly-selected customers purchase. Find E[I] and s.d.(I). (5 pts)
E[I] = E[12X - 6] = 12 E[X] - 6 = 12*1.7 - 6 = 14.4.
6
Var[I] = Var[12X - 6]= 122*Var[X]=144*0.81=116.64.
s.d.[I]= Var[ I ] = 144 * 0.81 = 12 * 0.9 = 10.8 .
7