Review for Midterm I

Study Guide
1. How to list the elements of the outcome (sample) space (and of possible events)
of a random experiment.
2. How to apply the three axioms of probability and the five theorems that followed.
3. Counting techniques -- multiplication rule, permutations (ordered) of n distinct
objects, permutation of n distinct objects taken r at a time, combinations
(unordered), distinguishable permutations of objects that are not all distinct.
4. Definition of conditional probability and the multiplication rule.
5. Definitions of independent events.
6. How to check to see if a function f(x) is a valid probability mass function.
7. How to calculate the expectation of any function of a discrete random variable.
8. How to calculate the mean, variance and standard deviation of any discrete
random variable.
9. The rules for expectation, mean, and variance.
10. All material covered in class so far, and how to solve every assigned problem.

Practice Problems
1. Describe the outcome space for each of the following experiments and type of data:
(a) A rat is selected at random from a cage, and its sex is determined.
Outcome space: S={female, male}
Type of Data: Binary
(b) A state selects a three-digit integer at random for one of its lottery games.
Outcome space: S={000, 001, 002, 003, …, 999}
Type of Data: Discrete
2. If S=AUB, P(A)=0.7, and P(B)=0.9, find
(a) P(A∩B):
Since P(S)=P(AUB)=1,
1= P(AUB)=P(A)+P(B)-P(A∩B)=0.7+0.9-P(A∩B)
Therefore, P(A∩B)=1.6-1=0.6
(b) P(A’UB’):
Since P(A’UB’)=1-P(A∩B),
P(A’UB’)=1-P(A∩B)=1-0.6=0.4
3. A certain state uses license plates consisting of two letters of the alphabet (A,…,Z)
followed by four digits (0,….,9). What is the probability that a randomly selected individual
from the state is assigned a license plate with two different letters and four different
digits?
Since N(S)=26*26**10*10*10*10=6760000, and N(A)=26*25**10*9*8*7=3276000,
3276000
N ( A)

 0.48
N ( S ) 6760000
4. There are six men, two women, and two children traveling on a bus. Four people get off
the bus at the next stop. What is the probability that the people get off the bus consist of
three men and one child?
Since N(S)=10C4, and N(A)=6C3*2C0*2C1,
 6  2  2
  *   *  
3
0
1
40
N ( A)
     
 0.19
N (S )
210
10 
 
4
5. A family with three children is randomly selected. The gender of each child in the family is
noted. Assume that the genders are equally likely. If at least one of the children in the
selected family is a boy, what is the probability that at least 2 of the children in the family
are boys?
Let A={At least 2 of children in the family are boys} and B={At least 1 of children in
the family is boy}, then A∩B={At least 2 of children in the family are boys}. The
probability that at least 2 of the children in the family are boys if at least one of
the children in the selected family is a boy is
P( A | B) 
P(# of boys  2)  P(# of boys  3)
P( A  B)

P( B)
P(# of boys  1)  P(# of boys  2)  P(# of boys  3)
 3  1 
 3  1 
1
1
   (1  )1     (1  ) 0 4
2
2
P(# of boys  2)  P(# of boys  3)  2  2 
4
 3  2 


 8 
0
7
1  P(# of boys  0)
7
 3  1 
1
8
1     (1  ) 3
2
 0  2 
2
3
6. Suppose that P(A∩B)=0.2 and P(A)=0.6 and P(B)=0.5. Are A and B independent events?
Since P(A∩B)=0.2 and P(A)*P(B)=0.6*0.5=0.3, P(A∩B)≠P(A)*P(B). Thus, A and B are
not independent events.
7. An insurance company sells an automobile policy with a deductible of one unit. Let X be
the amount of the loss having pmf,
 0.9
f ( x)  
c / x
x0
x  1,2
where c is a constant.
(a) Determine c so that f(x) satisfies the conditions of being a pmf for a r.v. X
Using the definition of pmf that
f ( x)  1 ,

xR
0.9  c / 1  c / 2  1   c  1 / 15
(b) Find E(X), E(X2), and E(3X2-2X+4)
22 2

30
15
x0 ,1, 2
1
1
24 1
E ( X 2 )   x 2 f ( x)  0 * 0.9  12 *
 22 *


15
*
1
15
*
2
30
5
x0 ,1, 2
1
2
1
13
E (3 X 2  2 X  4)  3E ( X 2 )  2 E ( X )  4  3 *  2 *  4   4 
5
15
3
3
E( X ) 
1
1
 xf ( x)  0 * 0.9  1* 15 *1  2 * 15 * 2 
(c) Find the mean, Variance, and standard deviation of a r.v. X
Mean:   E ( X )  2
15
1
2
1
4
94
5
Variance:  2  E ( X 2 )  E ( X ) 2       



1
2
5
Standard deviation:    2  1
45
2
 15 
 1
25
225
225
225
45
3 5
8. In the gambling game, the player wins $1 with probability 0.49293 and loses $1 with
probability 0.50707 for each bet. What is the expected value of the game to the player?
0.49293 x  $1
,
0.50707 x  $1
Since f ( x)  
E( X ) 
 xf ( x) 0.49293 * $1  0.50707 * $1  $0.01414
x $1, $1
9. Given E(X+4)=10 and E[(X+4)2]=116, determine Var(X+4), E(X), and Var(X).
Var ( X  4)  E[( X  4) 2 ]  E ( X  4) 2  116  102  16
 Var ( X ) (Var ( X  a)  Var ( X ) )
Since E ( X  4)  E ( X )  4  10 ( E ( X  a)  E ( X )  a ) , E ( X )  6