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Chapter 4 Review
This review is not a complete review. You should look at
your homework and read your textbook.
As always make sure you understand the vocabulary and
the concepts in this chapter.
Chapter 4 review
1) The probability of a person not dying in a car accident in a
one year period is 0.992. What is the probability of a person
dying in a car accident?
P(person dying in car accident) = 1 - P(person not dying in car accident)
= 1 - .992
= 0.008
2) The probability of a person not dying in a car accident in a
one year period is 0.992. If I chose at random 3 people from
the population of drivers that the stated probability refers to,
what is the probability of all 3 people not dying in a one year
period? Assume the events are independent.
P(person not dying and person not dying and person not dying)=
P(person not dying) P(person not dying) P(person not dying)=
(.992) (.992) (.992)  .9762
3) The probability of a person not dying in a car accident in
a one year period is 0.992. If I chose at random 3 people
from the population of drivers that the stated probability
refers to, what is the probability of at least one person
dying?
P(at least one person) = 1 - P(no one dies)
= 1 - .9762
= .0238
Note: the probability, P(no one dies) was calculated in the previous problem.
4) Let the random variable X denote the number of correct
answers on an exam with 10 questions.
X
P(X)
0
0.01
1
0.01
2
0.02
3
0.02
4
5
6
0.1 0.15 0.19
7
0.3
8
0.18
A) Is this a legitimate probability distribution? Explain.
No, the sum of the probabilities does not equal 1.
9
10
0.05 0.02
4) Let the random variable X denote the number of correct
answers on an exam with 10 questions.
X
P(X)
0
0.01
1
0.01
2
0.02
3
0.02
4
5
6
0.1 0.15 0.19
7
0.25
8
0.18
9
10
0.05 0.02
The probability distribution table is now legitimate.
B) What is the probability of a student chosen at random to
get a score of a 3 or a 4 on this exam? Use appropriate
probability notation to write your answer.
P(x=3 or x = 4) = P(x=3) + P(x = 4)
= .02 + .1
= .12
4) Let the random variable X denote the number of correct
answers on an exam with 10 questions.
X
P(X)
0
0.01
1
0.01
2
0.02
3
0.02
4
5
6
0.1 0.15 0.19
7
0.25
8
0.18
9
10
0.05 0.02
C) If the number of correct responses that students achieve in
this exam are considered independent, what is the probability
that two students chosen at random get at least 5 problems
correct on this exam. Use appropriate probability notation to
write your answer.
P(X  5 for first student and X  5 for second student) =
P(X  5 ) P(X  5 ) = (.84)(.84)
= .7056
4) Let the random variable X denote the number of correct
answers on an exam with 10 questions.
X
P(X)
0
0.01
1
0.01
2
0.02
3
0.02
4
5
6
0.1 0.15 0.19
7
0.25
8
0.18
9
10
0.05 0.02
D) What is the probability of three students chosen at random
to get at least 7 answers correctly, or between 5 and 8 answers
correctly inclusive.
P(X  7or5  X  8) = P(X  7) +P(5  X  8)- P(X  7and5 X  8)
= 0.5 + 0.77 - (0.25+0.18)
= 0.84
5) A random number generator has been set to choose any
number between 10 and 20. The distribution is a uniform
distribution as shown below.
20 X
10
A. Calculate P(9 < X < 18)
= (1/10)(18 - 10) = 8/10
B. Calculate P(X<12 or X> 15) = (1/10)(12-10) +(1/10)(20-15)
= 2/10 + 5/10 = 7/10
C. Calculate P(X > 15 | X<18)
=[(1/10)(18-15)]/[(1/10)(18-10)]
= 3/8
6) Lets say that the probability of being born a girl is 0.55
and the odds of being born a boy are 0.45. Let the random
variable X count the number of girls for a family of four
children.
A. Write out the sample space for the random variable X.
X = {0, 1, 2, 3, 4}
B. Write the probability distribution for the random variable X.
X
0
1
2
3
4
P(x)
.041
.200
.368
.299
.092
7) Suppose you work for an insurance company, and you sell a
$10,000 1-year term insurance policy at an annual premium of
$290. Actuarial tables show that the probability of death during
the next year for a person of your customer’s age, sex, health,
etc., is .001. What is the expected gain (amount of money by the
company) for a policy of this type?
μ = (-$10,000 + $290)(0.001) + $290(1 - 0.001)
μ = $280
To answer the question I had to consider all the possible simple events that pertain
to the problem and multiply these by the corresponding probabilities.
8) Medical research has shown that a certain type of
chemotherapy is successful 70% of time when used to treat
skin cancer. Suppose five skin cancer patients are treated with
this type of chemotherapy and let the random variable, X,
equal the number of successful cures out of the five.
X
0
1
2
3
4
5
P(x) .002 .029 .132 .309 .360 .168
If five patients were treated at a time, what is the
average number of successful cures we would expect to
see out of five patients?
μ = 0(0.002) + 1(0.029) + 2(0.132) + 3(0.309) + 4(0.360) + 5(0.168)
9) The Hawks hockey team has a 30% chance of winning against
Blue Devils on the first of a two game series, a 60% of losing and
a 10% chance of tying.
If the Hawks win the first encounter the chance of winning the
second game is 30%, and a 20% chance of losing.
If the Hawks lose the first encounter, then the chance of winning
the second game is 50%, and the chance of losing is 40%
A. Write the probabilities given using probability notation,
P(A), P(A | B), P(A and B) and so on.
P(win first game) = 0.30 , P(losing first game) = 0.60
P(tying first game) = 0.10
P(win second game | win first game) = 0.30
P(lose the second game | win the first) = 0.20 and so on.
9) The Hawks hockey team has a 40% chance of winning against
Bluedevils on the first of a two game series, a 40% of losing and a
20% chance of tying.
If the Hawks win the first encounter the chance of winning the second game
goes down to 30%, and a 50% chance of losing.
If the Hawks lose the first encounter, then the chance of winning the second
game is 50%, and the chance of losing is 40%
B. Are the events “tying in the first game” and “tying in the
second game”, independent?
Yes since P(tying) = .20 and the
P(tying second game | tied first game) = .20 are the
same.
9) The Hawks hockey team has a 40% chance of winning against
Bluedevils on the first of a two game series, a 40% of losing and a
20% chance of tying.
If the Hawks win the first encounter the chance of winning the
second game goes down to 30%, and a 50% chance of losing.
If the Hawks lose the first encounter, then the chance of winning
the second game is 50%, and the chance of losing is 40%
C. Create a tree diagram of the situation presented.
P(W | W) = 0.30
W
W
P(L |W) = 0.50
L
P(T|W) = 0.20
T
P(W) = 0.40
P(W | L) = 0.50
P(L) =0.40
L
P(L | L) = 0.40
L
P(T | L)=0.10
P(T) =0.20
T
W
T
9) The Hawks hockey team has a 40% chance of winning against
Bluedevils on the first of a two game series, a 40% of losing and a
20% chance of tying.
If the Hawks win the first encounter the chance of winning
the second game goes down to 30%, and a 50% chance of
losing.
If the Hawks lose the first encounter, then the chance of
winning the second game is 50%, and the chance of losing is
40%
E. What is the probability that the
Hawks will win the first game and
lose the second game?
P(W | W) = 0.30
W
W
P(L |W) = 0.50
L P(W and L) = (0.40)(0.50)
P(T|W) = 0.20
T
P(W) = 0.40
P(W | L) = 0.50
P(L) =0.40
L
P(L | L) = 0.40
L
P(T | L)=0.10
P(T) =0.20
T
W
T
9) The Hawks hockey team has a 40% chance of winning against
Bluedevils on the first of a two game series, a 40% of losing and a
20% chance of tying.
If the Hawks win the first encounter the chance of winning
the second game goes down to 30%, and a 50% chance of
losing.
If the Hawks lose the first encounter, then the chance of
winning the second game is 50%, and the chance of losing is
40%
F. What is the probability that the
Hawks will win one game?
P(W | W) = 0.30
W
W
P(L |W) = 0.50
L
P(T|W) = 0.20
T
P(W and L) = (0.40)(0.50)
P(W) = 0.40
P(W | L) = 0.50
P(L) =0.40
P(L | L) = 0.40
L
L
P(T | L)=0.10
P(T) =0.20
T
P(win one game) = P( W and L OR
L and W)
= (0.40)(0.50) + (0.40)(0.50)
= 0.40
W
T
P(L and W) = (0.40)(0.50)
10) Given P(A) = 0.2; P(B) = 0.4; P(A or B) = 0.5.
A. Calculate P(A and B).
P(A) + P(B) - P(A or B) = 0.2 + 0.4 - 0.5
= 0.1
10) Given P(A) = 0.2; P(B) = 0.4; P(A or B) = 0.5.
B. Calculate P(Ac or Bc).
Everywhere the lines
extend to is an area
represented by Ac or Bc.
A
B
P(Ac or Bc) = 1 - P(A and B)
= 1 - 0.10
= 0.90
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