Kansrekening­Engelse versie
Een toets Kansrekening en Statistiek
Exercise 1. Sarah is playing a game of cards with her friends. She is about to draw a card from the deck of
cards. She will win the game if at least one of the following two events happens. A : she draws an
ace; B : she draws "hearts". Otherwise she will lose the game. How can the event that Sarah will
win be expressed in terms of A and B?
a. A
b B.
b. A
c B.
c. A +B
a
b
.
c
Elaboration: the event that at least one of two events happens is denoted as A c
sets).
B
(the union of two
Exercise 2. Four components are connected to form a system as shown in the following diagram. The whole
system will only function if there is a path/connection of working components from left to right.
Denote by A i the event that i­th component works. What is the event that the system functions?
a. (A 1
bA
b. (A 1
c
c. (A 1
bA
d. (A 1
c
a
b
2)
A2 )
2)
A2 )
c
b (A b A ).
3
4
c
(A 3
c
c
(A 3
b A ).
b (A c
3
d
A4 )
.
4
A4 )
.
Elaboration: These are two serially connected subsystems, say S 1 and S 2 , each with two
components connected in parallel. So the event that the system functions is S 1 b S 2 . Since S 1 and S 2 have two components in parallel, the event that the system functions is ( A 1 c A 2 ) b ( A 3 c A 4 ) .
Exercise 3. Four components are connected to form a system as shown in the following diagram. The whole
system will function if there is a path/connection of working devices from left to right.
Denote by A i the event that i­th component works. What is the event that the system functions?
a. (A 1
bA
b. (A 1
c
c. (A 1
bA
d. (A 1
c
a
b
2)
A2 )
2)
A2 )
c
b (A b A ).
3
4
c
(A 3
c
c
(A 3
b A ).
b (A c
3
A4 )
.
4
A4 )
.
d
Elaboration: These are two parallel subsystems, say S 1 and S 2 , each with two components
connected serially. So the event that the system functions is S 1 c S 2 . Since S 1 and S 2 have two
components connected serially, the event that the system functions is ( A 1 b A 2 ) c ( A 3 b A 4 ) .
Exercise 4. Three letters, for Anthony, Betty and Carla are put ­ without looking ­ into three envelopes already
addressed to Anthony, Betty and Carla. What is the probability that exactly one envelope will
contain the correct letter?
a. 1/3
.
b. 1/2
.
c. 2/3
.
a
b
c
Elaboration: There are 6 elements in the sample space. They have the same probability. Three of the
1/2
6 outcomes satisfy the criterion that exactly one gets the letter correctly. So, the probability is 1/2.
Exercise 5. The probability that a randomly selected student at a certain university has a Visa credit card is 0.5.
The probability that the student has a Mastercard is 0.4. Furthermore the probability that the student
has both types of cards is 0.25. What is the probability that a randomly selected student from the
university has at least one credit card?
a. 0.9
b. 1.15
.
c. 0.65
.
a
.
b
c
Elaboration: Using the addition rule, one finds P(A c
B) = P(A) + P(B) − P(A
b B)
= 0.65
Exercise 6. For the two events A and B it is known that P( A
c
bB
c
) = 1/2
and that P(A)
= P(B) = 1/3
.
What is P(A b B) ?
a. 1/2
b. 1/9
c. 1/6
a
b
c
Elaboration: P(A
b B)
Using = P(A) + P(B) − P(A
c
the addition rule, B) = P(A) + P(B) − (1 − P( A
c
bB
one c
finds )) = 1/3 + 1/3 − (1 − 1/2) = 1/6
Exercise 7. When the electricity usage of a household falls below a certain kWh in a month, the electricity
supplier offers a special rate to the household. Let A denote the event that usage of electricity of a
certain household in the month of August is below the threshold. Let B denote the event that the
electricity usage of the same household in the month of December is below the threshold. Suppose
P(A) = 0.4, P(B) = 0.3
and P(A c
B) = 0.6
. What is the probability that the electricity usage
of the household would be below the threshold in both August and December?
a. 0.12
b. 0.7
.
.
.
c. 0.1
a
.
b
c
Elaboration: P(A
b B)
Using = P(A) + P(B) − P(A
c
the addition B) = 0.4 + 0.3 − 0.6 = 0.1
rule, .
one finds Exercise 8. The following table summarizes what happened to the 2223 passengers of the sinking Titanic ship.
Men Women Boys Girls
Survived 332 318
29
27
If one randomly selects a passenger from the passengers list, what is the probability of getting a
man or a child?
a. 0.8102
b. 0.7899
c. 0.5496
a
b
c
Elaboration: The number of boys, girls and men is 1801. So, the probability is 1801/2223.
Exercise 9. A bus manufacturing company buys mechanical parts to be built into the bus doors from many
smaller suppliers. Due to faulty parts coming from one of the suppliers, the doors of some buses
start malfunctioning shortly after the buses are put into service. A certain city has bought 20 such
buses from this company, out of which 8 have door parts from the faulty supplier. The city council
wants to make a thorough inspection of these buses. It selects the buses one by one at random
(and, of course, without replacement). What is the probability that the second bus chosen has faulty
door parts given that the first selected bus has faulty door parts?
14
a. 95
7
b. 19
8
c. 19
8
d. 20
a
b
c
d
Elaboration: This is a conditional probability. If the first bus was 'faulty', only seven 'faulty buses' remain
19
7/19
out of 19. The probability is 7/19.
Exercise 10. A bus manufacturing company buys mechanical parts to be built into the bus doors from many
smaller suppliers. Due to faulty parts coming from one of the suppliers, the doors of some buses
start malfunctioning shortly after the buses are put into service. A certain city has bought 20 such
buses from this company, out of which 8 have door parts from the faulty supplier. The city council
wants to make a thorough inspection of these buses. It selects the buses one by one at random
(and, of course, without replacement). What is the probability that both of the first two chosen buses
have faulty door parts?
14
a. 95
7
b. 19
c. (
a
8
20
b
)
2
c
Elaboration: Use the multiplication rule P(A b B)
to 8/20 ˝7/19 = 14/95.
\ to find that the probability is equal
= P(A)P(B A)
Exercise 11. The following table summarizes what happened to the 2223 passengers of the sinking Titanic ship.
Men Women Boys Girls
Survived 332 318
29
27
One of the Titanic passengers is randomly selected from the list who started the journey. If the
selected passenger is a man, what is the probability that he survived the disaster?
a. 0.7611
b. 0.1493
c. 0.4703
d. 0.1962
a
b
c
d
Elaboration: this is a conditional probability. The answer is
332/2223
332
=
1692/2223
Exercise 12. = 0.1962.
1692
A fire insurance company has high­risk (H), medium­risk (M) and low­risk (L) clients, who have,
respectively, probabilities 0.02, 0.01, 0.0025 of filing claims (C) within a given year. The
proportions of clients in these three categories are 0.1, 0.2, and 0.7, respectively. What is the
probability that a randomly chosen client will file a claim in a given year?
a. 0.0058
b. 0.0325
c. 0.0108
d. 0.0014
a
b
c
d
Elaboration: Use the 'law of total probability' to find \begin{equation*}
\
\
\
P(C) = P(C H)P(H) + P(C M)P(M) + P(C L)P(L) =
= 0.02
˝0.1 + 0.01 ˝0.2 + 0.0025 ˝0.7
= 0.0058.
\end{equation*}
Exercise 13. Four components are connected to form a system as shown in the following diagram. The whole
system will function if there is a path/connection of working devices from left to right.
Suppose the components work independently of one another and each one of them has 90%
chance of working. What is the probability that the system functions?
4
a. (0.9)
b. (0.99)
c. 0.81 + 0.81 − (0.81)
d. 0.99 + 0.99 − (0.99)
a
b
.
2
c
2
.
2
.
d
Elaboration: These are two serially connected subsystems, say S1
and S1
S2
, each with two
bS
2
S1
components connected in parallel. So the event that the system functions is S 1 b S 2 . Since S 1 and S 2 have two components in parallel, the event that the system functions is ( A 1 c A 2 ) b ( A 3 c A 4 ) .
Therefore, the probability is
P[( A 1
since P( A 1
c
c
A2 )
b (A c
3
A 2 ) = 1 − (0.1 )
2
A 4 )] = P( A 1
= 0.99
c
A2 )
˝P( A c
3
A 4 ) = (0.99)
2
,
.
Exercise 14. Four components are connected to form a system as shown in the following diagram. The whole
system will function if there is a path/connection of working devices from left to right.
Suppose the components work independently of one another and each one of them has 90%
chance of working. What is the probability that the system functions?
a. 0.81 + 0.81 − (0.81)
b. 0.99 + 0.99 − (0.99)
c. (0.9)
d. (0.99)
a
b
4
2
.
2
.
.
2
.
c
d
Elaboration: These are two parallel subsystems, say S 1 and S 2 , each with two components
connected serially. So the event that the system functions is S 1 c S 2 . Since S 1 and S 2 have two
components connected serially, the event that the system functions is ( A 1 b A 2 ) c ( A 3 b A 4 ) .
Therefore, the probability is \begin{equation*}
P[( A 1
bA
2)
c
(A 3
bA
4 )]
= 0.81 + 0.81 − (0.81)
\end{equation*} since P( A 1
bA
2)
2
= P( A 1
bA
2)
+ P( A 1
bA
2)
− P[( A 1
bA
2)
b (A b A
3
,
= 0.81
.
Exercise 15. A fire insurance company has high­risk (H), medium­risk (M) and low­risk (L) clients, who have,
respectively, probabilities 0.02, 0.01, 0.0025 of filing claims (C) within a given year. The
proportions of clients in these three categories are 0.1, 0.2, and 0.7, respectively. What proportion
of the claims filed each year come from high­risk clients?
4 )]
a. 0.1
b. 0.02
c. 0.3478
d. 0.002
a
b
c
d
Elaboration: the probability is P(H\C) . Using Bayes' theorem one finds
\
P(H)
\
P(H C) = P(C H)
= 0.02
P(C)
˝
0.1
= 0.3478,
0.00575
since \begin{equation*}
\
\
\
˝0.1 + 0.01 ˝0.2 + 0.0025 ˝0.7
P(C) = P(C H)P(H) + P(C M)P(M) + P(C L)P(L) =
= 0.02
= 0.00575.
\end{equation*}
Exercise 16. Suppose 60% of a company's computer chips are made in factory A and 40% are made in factory B
. Furthermore, the rate of defective chips produced by factory A is 30% and that for factory B is
20%. If a randomly chosen chip in the company's computer turned out to be defective, what is the
probability that the chip came from factory A?
a. 0.18
b. 0.6923
c. 0.30
d. 0.5
a
b
c
d
Elaboration: the probability is P(A\D) . Using Bayes' theorem one finds
\
\
P(A)
P(A D) = P(D A)
= 0.3
P(D)
˝
0.6
= 0.6923,
0.26
since \begin{equation*}
\
\
= 0.3 ˝0.6 + 0.2 ˝0.4 = 0.26.
P(D) = P(D A)P(A) + P(D B)P(B)
\end{equation*}
Exercise 17. Two events A and B are mutually exclusive. We have P(A)
Give P(A b B
c
)
= 0.26`
and P(B)
= 0.22`
.
.
Elaboration: because c
P(A b B ) = P(A) =.
A
and B
are mutually exclusive, we have A
à B . Therefore, c
Exercise 18. For two events A and C the following holds:
, P(A) = 0.25 P(C) = 0.35
and P(A b C)
= 0.1
.
Compute P(A b C ) .
c
Elaboration: with the total probability rule one finds P(A)
c
P(A b C ) = 0.25 − 0.1 = 0.15 .
= P(A
b C) + P(A b C ) . Therefore, c
Exercise 19. Consider the following circuit.
This circuit operates if there is a path of functional devices from left to right. There are three
devices (I, II and III). The probability that each device functions is given in the graph. Assume that
devices fail independently.
It is given that the circuit operates. Given this fact, what is the probability that device I functions?
Elaboration: this is conditional probability P(A \ \makeboxcircuit operates) , where A denotes the
event that device I functions. Use the definition of conditional probability to evaluate this probability or
use Bayes' theorem to evaluate it. With Bayes' theorem one finds \begin{equation*}
\
P(A \makeboxcircuit operates) =
\
P(\makeboxcircuit operates A)
˝P(A)
\
P(\makeboxcircuit operates A)
\
P(A \makeboxcircuit operates) =
˝0.90) ˝0.85
0.90 ˝0.85 + 0.70 − 0.70 ˝0.90 ˝0.85
(0.90 + 0.70 − 0.70
=
˝P(A)
P(\makeboxcircuit operates)
0.8245
=
= 0.887
0.9295
\end{equation*}
Exercise 20. Two events A and B are mutually exclusive. We have P(A)
For another event C we have P(C)
Furthermore P(A b C)
Compute P(A c
B
c
= P(B
C)
= 0.35
b C)
= 0.25
and P(B)
= 0.4
.
.
= 0.1
.
.
Elaboration: Use the addition rule for three events to find \begin{equation*}
P(A
c
B
c
C)
= P(A) + P(B) + P(C) − P(A
b B) − P(A b C) − P(B b C) + P(A b B b C)
0.25 + 0.4 + 0.35 − 0 − 0.1 − 0.1 − 0 = 0.8.
\end{equation*}
Exercise 21. A meeting has been planned for the board of directors of a company. It is essential that both the
CEO (Roy) and the CFO (Inge) are present at the meeting. If one of them (or both) has to cancel,
the meeting will be canceled also. The probability that Roy has to cancel the meeting is 1/10, the
probability that Inge has to cancel the meeting is 1/5. The two events that Roy cancels the meeting
and that Inge cancels the meeting are independent.
The meeting has been canceled. Given this fact, what is the probability that Both Roy and Inge
canceled the meeting?
Elaboration: Let A denote the event that Roy cancels and B the event that Inge cancels. The
conditional probability is \begin{equation*}
P[(A
b B)\(A c
P[(A
B)]
=
b B) b (A c
P(A c B)
B)]
b B)
P(A c B)
P(A
=
P(A)P(B)
=
P(A) + P(B) − P(A
\end{equation*}
1/10
b B)
=
P
1/5
1/10 + 1/5 − 1/10
P
= 1/14.
1/5
Exercise 22. Two events A and B are mutually exclusive and we have
P(A) = 0.3,
Compute the probability P( A
c
P(B) = 0.4
b B ).
c
Elaboration: Using Morgan's law and the addition rule we have \begin{equation*}
P( A
c
bB
c
) = 1 − P(A
c
B)
= 1 − P(A) − P(B) + P(A
b B)
= 1 − 0.3 − 0.4 − 0 = 0.3.
\end{equation*}
Exercise 23. For three events A, B and C we have
P(A) = 0.3,
P(C
\ A)
P(B) = 0.4,
= 0.9,
P(C
\ B)
= 0.6,
P(C
\ (A b B
C
C
)) = 0.1 .
Furthermore the two events A and B are mutually exclusive.
Compute the probability P(C) .
Elaboration: Because A c B c ( A b B c ) = S and A, B and ( A
can use the total probability rule to find \begin{equation*}
c
\
\
0.9 ˝0.3 + 0.6 ˝0.4 + 0.1 ˝0.3
c
P(C) = P(C A)P(A) + P(C B)P(B) + P(C
=
\end{equation*} b B ) are mutually exclusive one
c
\ (A b B
′
= 0.54.
′
))P(A
′
bB
′
) =