MAT 326
Probability and statistics for
Engineers
Chapter 4: Probability
4.1: Sample Space and relationships among
events
• A sample space is a set of all possible outcomes of an experiment.
• Notation: S
• An event is any subset of a sample space.
• Notation: A, B, C
Examples
• Sample space associated with a die toss: S={1,2,3,4,5,6}
• Consider the event A: observing an even outcome. A={2,4,6}
• Sample space associated with Flipping a coin: S={T,H}
• T: tail,
H: head
Examples
• Flip two coins: S={HH,HT,TH,TT}
• B: observing at least one head; B={HH,HT,TH}
• C: Observing at most one tail; C={HH,HT,TH}
• D: Observing exactly one head; D={HT,TH}
Examples
• Roll two dice: S={(1,1),(1,2),…,(1,6),(2,1),(2,2),…,(6,6)}
• A: sum of outcomes is equal to 11; A={(5,6),(6,5)}
• B: sum of outcomes is at least 11; B={(5,6),(6,5),(6,6)}
Relationships among events
• Let 𝑆 be a sample space from an experiment and A and B two events.
• Union: 𝐴 ∪ 𝐵 = {𝑥|𝑥 ∈ 𝐴 𝑜𝑟 𝑥 ∈ 𝐵}
• Intersection: 𝐴 ∩ 𝐵 = {𝑥|𝑥 ∈ 𝐴 𝑎𝑛𝑑 𝑥 ∈ 𝐵}
• Complement of an event: 𝐴ҧ = {𝑥 ∈ 𝑆 𝑎𝑛𝑑 𝑥 ∉ 𝐴}
Relationships among events
• Mutually exclusive events: 𝐴 ∩ 𝐵 = ∅ (empty set)
• 𝐴 ∪ 𝐴ҧ = 𝑆
• De Morgan’s laws:
• 𝐴 ∪ 𝐵 = 𝐴ҧ ∩ 𝐵ത
• 𝐴 ∩ 𝐵 = 𝐴ҧ ∪ 𝐵ത
Venn diagram
• Mutually exclusive events
B
A
Venn diagram
• Complement of an event
Venn diagram
• Union of two events
Venn diagram
• Intersection of 2 events
Example
• Flip a coin 3 times: S={HHT,HTH,HTT,HHH,THH,THT,TTH,TTT}
• A: at least one head; A={HHT,HTH,HTT,HHH,THH,THT,TTH}
• B: exactly one head; B={HTT,THT,TTH}
• 𝐴∩𝐵 =𝐵
• 𝐴∪𝐵 =𝐴
• 𝐴:ҧ No head; 𝐴ҧ = {𝑇𝑇𝑇}
• 𝐴 ∪ 𝐴ҧ = 𝑆
• 𝐴ҧ ∩ 𝐵 = ∅
H
T
H
H
T
T
H
T
H
H
T
H
T
T
4.2-Definition of probability
• Suppose S is a sample space associated with an experiment. A
probability is a numerically valued function that assigns a number
p(A) to every event A such that
•𝑝 𝐴 ≥0
•𝑝 𝑆 =1
• If 𝐴1 , 𝐴2 , … , 𝐴𝑛 is a sequence of mutually exclusive events, then
• 𝑝 𝐴1 ∪ 𝐴2 ∪ ⋯ ∪ 𝐴𝑛 = σ𝑛𝑖=1 𝑝 𝐴𝑖
Equally likely outcomes
• Roll a die. S={1,2,3,4,5,6}
• Probability of each outcome is equal to 1/6
• Roll two dice: S={(1,1),(1,2),…,(6,6)}
• Probability of each outcome is equal to 1/36
Equally likely outcomes
• In general, if S is a finite sample space of some experiment and A is
any event, under the assumption of equally likely outcomes, the
probability that event A occurs is given by
•𝑝 𝐴 =
𝐴
𝑆
• 𝑆 denotes the cardinality of the set S, i.e., the number of elements
Examples
• Flip three coins: S={HHT,HTH,HTT,HHH,THH,THT,TTH,TTT}
• What’s the probability of obtaining at least one head.
• A: at least one head; A={HHT,HTH,HTT,HHH,THH,THT,TTH}
• p=7/8
• Roll two dice. What’s the probability that the sum of the outcomes is
8.
• A: sum of outcomes is 8: {(5,3),(3,5),(4,4),(6,2),(2,6)}
• p=5/36
Probability of the union of two events
• Let S be the sample space of some experiment. Suppose A and B are
two given events.
• 𝑝 𝐴∪𝐵 =𝑝 𝐴 +𝑝 𝐵 −𝑝 𝐴∩𝐵
• If the events A and B are mutually exclusive then
• 𝑝 𝐴∪𝐵 =𝑝 𝐴 +𝑝 𝐵
Example
• Consider the national percentages for HIV-related risk groups (Science,
1992) given in Table 4.3. A firm hires a new worker after a national
advertising campaign.
• A) What is the probability that the worker falls in the “risky partner”
category?
• B) What is the probability that the worker is in at least one of the risk
groups?
• C) If the firm hires 1,000 workers, how many are expected to be at risk if
the 1,000 came from the population at large?
• D) If the firm hires 1,000 workers, how many are expected to be at risk if
the 1,000 come from high-risk cities?
Risk Group
• Multiple partners
• Risky partner
• Transfusion recipient
• Multiple partner and
risky partner
• Multiple partner and
transfusion recipient
• Risky partner and
transfusion recipient
• All others
• No risk
national
Percent
Number
7.0
170
3.2
76
2.3
55
high risk cities
Percent Number
9.5
651
3.7
258
2.1
144
1.7
41
3.0
209
0.0
1
0.3
20
0.2
0.7
84.9
4
1
2,045
0.3
0.7
80.4
19
51
5,539
• A) Solutions to practical problems of this type always involve assumptions.
To answer parts (a) and (b) with the data given, we must assume that the
new worker is randomly selected from the national population, which
implies P(risky partner) = 0.032
• b Let’s label the seven risk groups, in the order listed in the table, E1, E2, ,
E7 Then the event of being in at least one of the groups can be written as
the union of these seven, that is,
E1∪ E2 ∪ … E7
• The event “at least one” is the same as the event “E1 or E2 or E3 or … or
E7” Because these seven groups are listed in mutually exclusive fashion,
P(E1∪ 𝐸2 ∪ 𝐸3 … ∪ 𝐸7) = P(E1) + P(E2) + ⋯ P(E7)
=0.070+0.032+0.023+0.017+0.0+0.002+0.007=0.151
• Note that the mutually exclusive property is essential here; otherwise, we
could not simply add the probabilities
• C) We assume that all 1,000 new hires are randomly selected from
the national population (or a subpopulation of the same makeup).
Then 15.1% of the 1,000, or 151 workers, are expected to be at risk.
• D) We assume that the 1,000 workers are randomly selected from
high-risk cities. Then 19.6%, or 196 workers, are expected to be at risk
4.3-Counting techniques
• For many problems, it’s not straightforward to count the possible
outcomes.
• Roll a dice 5 times
• All possible outcomes in a lottery.
• All possible phone numbers in a region.
• All possible car plates in a region.
Multiplication rule
• If 𝐸1 is an experiment with 𝑛1 outcomes and 𝐸2 is an experiment with
𝑛2 outcomes, then the experiment which consists of performing
𝐸1 first and then 𝐸2 consists of 𝑛1 ∙ 𝑛2 outcomes.
• Generalization for an arbitrary number of experiments.
• If 𝐸1 is an experiment with 𝑛1 outcomes, 𝐸2 is an experiment with
𝑛2 outcomes, and so on up to 𝐸𝑟 which is an experiment with 𝑛𝑟
outcomes, then the experiment which consists of performing 𝐸1 first
and then 𝐸2 up to 𝐸𝑟 consists of 𝑛1 ∙ 𝑛2 ∙ ⋯ ∙ 𝑛𝑟 outcomes.
Example
• How many different license plates are possible if Kentucky uses three
letters followed by three digits.
• 𝐸1 : choosing first letter, 𝐸2 : choosing second letter, 𝐸3 : choosing third
letter
• 𝐹1 : choosing first digit, 𝐹2 : choosing second digit, 𝐹3 : choosing third digit
• 263 ∙ 103
Permutations
• The number of arrangements or permutations of r objects taken from
n distinct objects is given by
•
𝑃𝑟𝑛
=
𝑛!
𝑛−𝑟 !
• 0! = 1, 𝑛! = 1.2. … 𝑛
• Proof: By the multiplication rule, The number of arrangements is
𝑛!
given by 𝑛. 𝑛 − 1 . 𝑛 − 2 … 𝑛 − 𝑟 − 1 =
𝑛−𝑟 !
Example
• From among 10 employees, three are to be selected for travel to three outof-town plants, A, B, and C, with one employee traveling to each plant.
Because the plants are in different cities, the order of assigning the
employees to the plants is an important consideration. The first person
selected might, for instance, go to plant A and the second to plant B. In
how many ways can the assignments be made?
• Solution : Because order is important, the number of possible distinct
assignments is
10!
𝑃310 = =10(9)(8)=720
7!
In other words, there are 10 choices for plant A but only nine for plant B
and eight for plant C. This gives a total of 10(9)(8) ways of assigning
employees to the plants
Example
• An assembly operation in a manufacturing plant involves four steps,
which can be performed in any order. If the manufacturer wishes to
experimentally compare the assembly times for each possible
ordering of the steps, how many orderings will the experiment
involve?
• Solution: The number of orderings is the permutation of n 4 things
taken r 4 at a time. (All steps must be accomplished each time.) This
turns out to be
4 4!
𝑃4 = =24
0!
because 0! 1 by definition.
Example
• Four names are drawn from the 24 members of a club for the offices
of President, Vice-President, Treasurer, and Secretary. In how many
different ways can this be done?
Combinations
• In permutation, order is important. But in many problems the order
of selection is not important and interest centers only on the set of r
objects.
• The number of subsets or combinations of size r that can be selected
from n different objects is given by
•
𝑛
𝑟
=
𝑛!
𝑟! 𝑛−𝑟 !
Example
• How many committees of two chemists and one physicist can be
formed from 4 chemists and 3 physicists?
• Number of ways of choosing two chemists:
• Number of ways of choosing one physicist:
• By multiplication rule, the answer is: 3.6=18
4
2
3
1
=
=
4!
2!2!
3!
1!2!
=6
=3
Exercise 14
• Two vehicles in succession are observed moving through the
intersection of two streets.
• A) List the possible outcomes, assuming each vehicle can go straight,
turn left, or turn right.
• B) Assuming the outcomes to be equally likely, find the probability
that at least one vehicle turns left.
• C) Find the probability that at most one vehicle turns.
• A) Sample space; S={SS,SL,SR,LS,LL,LR,RS,RL,RR}
• B) Let A be the event: at least one vehicle turns left. Then
A={SL,LS,LL,LR,RL}. Hence p=5/9
• C) Let B be the event: at most one vehicle makes a turn.
B={SS,SL,SR,LS,RS}. Hence p=5/9
Exercise 18
• Seven applicants have applied for two jobs. How many ways can the
jobs be filled if
• A) the first person chosen receives a higher salary than the second?
• B) There are no differences between the jobs?
• A)7.6=42 (permutations)
• B)
7
2
=
7!
2!5!
= 21 (combinations)
Exercise 19
• A package of six light bulbs contains two defective bulbs. If three
bulbs are selected for use, find the probability that none is defective.
• Sample space:
6
3
=
6!
3!3!
= 20
• Event: None of the three selected bulbs are defective:
• Probability:
4
3
=
4!
1!3!
=4
• The number of ways of partitioning 𝒏 distinct objects into 𝑘 groups
containing 𝑛1 , 𝑛2 , … 𝑛𝑘 objects, respectively, is
•
𝑛!
𝑛1 !𝑛2 !…𝑛𝑘 !
where
• σ𝑘𝑖=1 𝑛𝑖 = 𝑛
Exercise 21
• A flee of eight taxis is to be randomly assigned to three airports A, B,
and C, with two going to A, five to B, and one to C.
• A) In how many ways can this be done?
• B) What is the probability that the specific cab driven by Jones is
assigned to airport C?
• In how many ways can this be done?
8!
2!5!1!
=
6.7.8
2
= 168
• B) What is the probability that the specific cab driven by Jones is
assigned to airport C?
Exercises
1. How many different words can be created by rearranging the letters
in SELFIESTICK?
2. A multiple choice test has 5 questions and each question has 4
choices. How many ways are there to answer the questions?
What’s the probability that a student get all answers correctly?
• 1)
11!
2!2!2!
• 2) 45
• 1/45
• 3)A manufacturing company has two retail outlets. It is known that
30% of the potential customers buy products from outlet I alone, 50%
buy from outlet II alone, 10% buy from both I and II, and 10% buy
from neither. Let A denote the event that a potential customer,
randomly chosen, buys from I and B denote the event that the
customer buys from II. Find the following probabilities.
• p(A), p(𝐴 ∪ 𝐵), p 𝐴 ∩ 𝐵 , 𝑝 𝐵ത , 𝑝 𝐴 ∪ 𝐵 ,
3)p(A)=0.4,
• p(𝐴 ∪ 𝐵)=0.9,
• p 𝐴 ∩ 𝐵 = 0.1,
• 𝑝 𝐵ത = 0.4,
• 𝑝 𝐴 ∪ 𝐵 = 0.1,
• 4)Suppose that three employees are to be selected from ten to visit a
new plant. In how many ways can the selection be made?
• If two of the ten employees are female and eight are male, what is
the probability that exactly one female gets selected among the
three?
•
10
3
• 5)A firm is placing three orders for supplies among five different
distributers. Each order is randomly assigned to one of the
distributors, and a distributor can receive multiple orders. Find the
probability that all orders go to different distributors.
•
5.4.3
53
Conditional probability
• 𝑝 𝐴|𝐵 =
𝑝 𝐴∩𝐵
• 𝑝 𝐵|𝐴 =
𝑝 𝐴∩𝐵
• 𝑝 𝐴|𝐵 =
𝑝 𝐵|𝐴 𝑝 𝐴
𝑝 𝐵
𝑝 𝐴
𝑝 𝐵
Examples
• If we randomly pick two television sets in succession from a shipment
of 240 television sets of which 15 are defective, what is the
probability that they will be both defective?
• A: first television picked was defective
• B: second tv picked was defective
• 𝐴 ∩ 𝐵: both tv picked were defective
• 𝑝 𝐴 ∩ 𝐵 = 𝑝 𝐴 𝑝 𝐵|𝐴 =
15
240
∙
14
239
=
7
1912
Examples
• A drawer contains 4 black, 6 brown, and 8 red socks. Two socks are
selected at random from the drawer.
• A) What is the probability that both socks are of the same color?
• B) What is the probability that both socks are red if it is known that
they are of the same color?
• There are
18
2
=
18!
2!16!
= 153 ways to select two socks
• Let A be the event that two socks selected at random are of the same
color. Then the cardinality of A is given by 42 + 62 + 82 = 6 +
15 + 28 = 49
• Hence p(A)=49/153
• B: two socks selected are red. The cardinality of B is
• P(B|A)=p(A∩B)/p(A)=p(B)/p(A)=
28/153
49/153
=28/49=4/7
8
2
= 28
Independent events
• Two events A and B are independent iff p(A|B)=p(A)
• 𝑝 𝐴|𝐵 =
𝑝 𝐴∩𝐵
𝑝 𝐵
• This is equivalent to 𝑝 𝐴 ∩ 𝐵 = 𝑝 𝐴 𝑝 𝐵 .
• EX33) Suppose that p(A)=0.6, p(B)=0.3, and 𝑝 𝐴 ∩ 𝐵 =0.15. Are
events A and B independent? No.
Example: Classification of manufactured pens
Assembly line 1
Assembly line 2
Total
Defective-trash
8
2
10
Defective-to be fixed
13
27
40
Non defective
59
91
150
Total
80
120
200
• Suppose a pen is selected at random. What is the probability that it is a
defective-trash pen manufactured by assembly line 1?
• Suppose a pen selected at random was manufactured by assembly line 1.
then what is the probability that it is defective-trash?
• Are the events, defective-trash and produced by assembly line 2 dependent
or independent?
• Suppose a pen is selected at random. What is the probability that it is a
defective-trash pen manufactured by assembly line 1?8/200
• Suppose a pen selected at random was manufactured by assembly line 1.
then what is the probability that it is defective-trash?8/80
• Are the events, defective-trash and produced by assembly line 2 dependent
or independent? dependent
• P(defective trash)=10/200, p(produced by assembly line 2)=120/200
• P(defective trash manufactured by assembly line 2)=2/200
• 10/200x120/200 is different from 2/200.
Probability of complement of an event
• If 𝐴ҧ is the complement of an event A in a sample space S, then
𝑝 𝐴ҧ = 1 − 𝑝(𝐴).
Example
• A quality-control inspector has ten assembly lines from which to choose
products for testing. Each morning of a five-day week, she randomly selects
one of the lines to work on for the day. Find the probability that a line is
chosen more than once during the week
• Solution: It is easier here to think in terms of complements and first find
the probability that no line is chosen more than once. If no line is repeated,
five different lines must be chosen on successive days, which can be done
in ways. The total number of possible outcomes for the selection of five
lines without restriction is , by an extension of the multiplication rule.
Thus, P(no line is chosen more than once)=
P(a line is chosen more than once) =1-0.3=0.7
Exercise
Solution
• Design A: combine relays 1 and 2 into the equivalent relay 5 which
• opens with prob = 0.12 and closes with prob =1 − 0.12. Similarly,
• combine the relays 3 and 4 into the equivalent relay 6 which opens
• with prob= 0.12 and closes with prob = 1 − 0.12. The new circuit is
in series. Hence, the prob that the current will flow is 1 − 0.12 2 =
0.9801
• Design B: combine relays 1 and 3 into the equivalent relay 5 which
closes with prob = 0.92 and open with =prob 1 − 0.92. Similarly,
combine relays 2 and 4 into the equivalent relay 6 which closes with
prob = 0.92 and open with prob = 1 − 0.92. The new circuit is in
parallel. Hence, the prob that the current will flow is 1 − (1 −
0.92 )2 = 0.9639
Bayes’ rule
• If 𝐵1 , 𝐵2 form a partition of S (their union is equal to S and they are
mutually exclusive), and A is any event in S, then
• 𝑝 𝐵1 |𝐴 =
𝑝 𝐵1 𝑝 𝐴|𝐵1
𝑝 𝐵1 𝑝 𝐴|𝐵1 +𝑝 𝐵2 𝑝 𝐴|𝐵2
• Proof:
• 𝑝 𝐵1 |𝐴 =
𝑝 𝐴∩𝐵1
𝑝 𝐴
=
𝑝 𝐴∩𝐵1
𝑝 𝐴∩𝐵1 +𝑝 𝐴∩𝐵2
=
𝑝 𝐵1 𝑝 𝐴|𝐵1
𝑝 𝐵1 𝑝 𝐴|𝐵1 +𝑝 𝐵2 𝑝 𝐴|𝐵2
• If 𝐵1 , 𝐵2 , … , 𝐵𝑘 form a partition of S (their union is equal to S and
they are mutually exclusive), and A is any event in S, then
• 𝑝 𝐵1 |𝐴 =
𝑝 𝐵1 𝑝 𝐴|𝐵1
𝑝 𝐵1 𝑝 𝐴|𝐵1 +𝑝 𝐵2 𝑝 𝐴|𝐵2 +⋯+𝑝 𝐵𝑘 𝑝 𝐴|𝐵𝑘
28 p. 185
• Electric motors coming off two assembly lines are pooled for storage
in a common stockroom, and the room contains an equal number of
motors from each line. Motors are periodically sampled from that
room and tested. It is known that 10% of the motors from line I are
defective and 15% of the motors from line II are defective.
• i) If a motor is randomly selected from the stockroom and found to be
defective, find the probability that it came from line I.
• ii) Find the probability that a randomly selected motor is defective.
Solution
• i) 𝐵1 : motor came from line I.
• 𝐵2 : motor came from line II.
• A: motor is defective
• 𝑝 𝐵1 = 𝑝 𝐵2 = 0.5; 𝑝 𝐴|𝐵1 = 0.1; 𝑝 𝐴|𝐵2 = 0.15
• 𝑝 𝐵1 |𝐴 =
𝑝 𝐵1 𝑝 𝐴|𝐵1
𝑝 𝐵1 𝑝 𝐴|𝐵1 +𝑝 𝐵2 𝑝 𝐴|𝐵2
=
0.1
0.1+0.15
= 0.4
• ii) p(A)=𝑝 𝐵1 𝑝 𝐴|𝐵1 + 𝑝 𝐵2 𝑝 𝐴|𝐵2 =0.1+0.15=0.25
Nb 47 p.202
• Two methods A and B are available for teaching a certain industrial
skill. The failure rate is 20% for A and 10% for B. However, B is more
expensive and hence is used only 30% of the time. (A is used the
other 70%) A worker is taught the skill by one of the methods but fails
to learn it correctly. What is the probability that he was taught by
method A?
Solution
• A: worker is taught by method A
• B: worker is taught by method B
• C: Worker fails to learn
• 𝑝 𝐴|𝐶 =
𝑝 𝐴 𝑝 𝐶|𝐴
𝑝 𝐴 𝑝 𝐶|𝐴 +𝑝 𝐵 𝑝 𝐶|𝐵
=
0.7∙0.2
0.7∙0.2+0.1∙0.3
= 0.82
Nb 40 p.201
• An incoming lot of silicon wafers is to be inspected for defectives by
an engineer in a microchip manufacturing plant. In a tray containing
20 wafers, assume four are defective. Two wafers are to be randomly
selected for inspection. Find the probability that
• A) Both are nondefective
• B) At least one of the two is nondefective
• C) Both are nondefective given that at least one is nondefective
Solution
• Let A: Both are nondefective and B: at least one is nondefective
• A)
• B)
• C) 𝑝 𝐴|𝐵 =
𝑝 𝐵∩𝐴
𝑝 𝐵
=
𝑝 𝐴
𝑝 𝐵
Nb 37 p.200
• A purchasing office is to assign a contract for computer paper and a
contract for microcomputer disks to any one of the three firms
bidding for these contracts. (Any one firm could receive both
contracts.) Find the probability that
• A) Firm I receives a contract given that both contracts do not go to
the same firm.
• B) Firm I receives both contracts
• C) Firm I receives the contract for paper given that it does not receive
the contract for disks.
Solution
• Sample space: 9 outcomes
• 𝑃1 𝐷1 , 𝑃1 𝐷2, 𝑃1 𝐷3 , 𝑃2 𝐷1 , 𝑃2 𝐷2 , 𝑃2 𝐷3 , 𝑃3 𝐷1 , 𝑃3 𝐷2 , 𝑃3 𝐷3
• A: Firm I receives a contract
• B: Both contracts do not go to the same firm.
• A)
1
• B) p(Firm I receives both contracts)=
9
Continue
• C: Firm I receives the contract for paper
• D: Firm I does not receive the contract for disks
• C)
Ex
• Suppose A and B are independent events. Are 𝐴ҧ and 𝐵ത also
independent?
• 𝑝 𝐴ҧ ∩ 𝐵ത = 𝑝 𝐴 ∪ 𝐵 = 1 − 𝑝 𝐴 ∪ 𝐵 =
• 1− 𝑝 𝐴 +𝑝 𝐵 −𝑝 𝐴 ∩ 𝐵 = 1−𝑝 𝐴 −𝑝 𝐵 −𝑝 𝐴 𝑝 𝐵 =
1−𝑝 𝐴 −𝑝 𝐵 1−𝑝 𝐴 = 1−𝑝 𝐴 1−𝑝 𝐵
• = 𝑝 𝐴ҧ 𝑝 𝐵ത
• Hence 𝐴ҧ and 𝐵ത are also independent.
Ex
• Vehicles coming into an intersection can turn left or right or go
straight ahead. Two vehicles enter an intersection in succession. Find
the probability that at least one of the two vehicles turns left given
that at least one of the two vehicles turns.
• S={SS, SR,SL, RS,RR,RL, LS,LR,LL}
• 5/8
Ex
• Suppose a certain disease has an incidence rate of 0.1% (that is, it
afflicts 0.1% of the population). A test has been devised to detect this
disease. The test does not produce false negatives (that is, anyone
who has the disease will test positive for it), but the false positive rate
is 5% (that is, about 5% of people who take the test will test positive,
even though they do not have the disease). Suppose a randomly
selected person takes the test and tests positive. What is the
probability that this person actually has the disease?
• A: A person has the disease
• 𝐴:ҧ A person does not have the disease
• B: A person test positive
• 𝑝 𝐴|𝐵 =
𝑝 𝐴 𝑝 𝐵|𝐴
𝑝 𝐴 𝑝 𝐵|𝐴 +𝑝 𝐴ҧ 𝑝 𝐵|𝐴ҧ
=
0.001×1
0.001×1+0.999×0.05
= 0.0196
Ex 15
• A commercial building is designed with two entrances, door I and
door II. Two customers arrive and enter the building.
• A) List the elements of a sample space for this observational
experiment.
• B) If all elements in part (a) are equally likely, find the probability that
both customers use door I; then find the probability that both
customers use the same door.
• a. S {(C1d1, C2d1), (C1d1, C2d2), (C1d2, C2d1), (C1d2, C2d2)}
• b. P(both customers use door I) =1/4
P(both customers use the same door)=1/2
Example
• Five motors (numbered 1 through 5) are available for use, and motor
2 is defective. Motors 1 and 2 come from supplier I, and motors 3, 4,
and 5 come from supplier II. Suppose two motors are randomly
selected for use on a particular day.
• Let A denote the event that the defective motor is selected and B the
event that at least one motor comes from supplier I.
• Find P(A) and P(B)
First motor selected
Second motor selected
1
2
3
4
5
2
1
3
4
5
3
1
2
4
5
4
1
2
3
5
5
1
2
3
4
• The tree diagram shows that there are 20 possible outcomes for this
experiment, which agrees with our calculation using the product rule.
That is, there are 20 events of the form {1, 2}, {1, 3}, and so forth.
Because the motors are randomly selected, each of the 20 outcomes
has probability 1/20. Thus,
𝑃 𝐴
= 𝑃 1,2 ∪ 𝑃 2,1 ∪ 𝑃 2,3 ∪ 𝑃 2,4 ∪ 𝑃 2,5 ∪ 𝑃 3,2 ∪ 𝑃 4,2
• Similarly, we can show that B contains 14 of the 20 outcomes and that
P(B)=14/20=0.7
Example
• Five applicants for a job are ranked according to ability, with applicant
number 1 being the best, number 2 second best, and so on. These
rankings are unknown to an employer, who simply hires two
applicants at random.
• What is the probability that this employer hires exactly one of the two
best applicants?
• The number of possible outcomes for the process of selecting two
applicants from five is
• If one of the two best is selected, the selection can be done in
=2
• The other selected applicant must come from among the three
lowest-ranking applicants, which can be done in
=3
• Thus, the event of interest (hiring one of the two best applicants) can
2×3
occur in
=0.6
10
Example
• A section of an electrical circuit has two relays in parallel, as shown in
the Figure. The relays operate independently, and when a switch is
thrown, each will close properly with probability only 0.8.
• If the relays are both open, find the probability that current will flow
from s to t when the switch is thrown
1
t
s
2
• Let O denote an open relay and C a closed relay. The four outcomes
for this experiment are shown by the following:
• E1 (O,O) E2(O,C) E3(C,O) E4(C,C)
• Because the relays operate independently, we can find the probability
for these outcomes as follows:
P(E4) = P(C)P(C) = (0.8)(0.8) = 0.64
P(E3) = P(C)P(O) = (0.8)(0.2) = 0.16
P(E2) = P(O)P(C) = (0.2)(0.8) = 0.16
P(E1) = P(O)P(O) = (0.2)(0.2) = 0.04
• If A denotes the event that current will flow from s to t, then
𝐴 = 𝐸2 ∪ 𝐸3 ∪ 𝐸4 𝑜𝑟 𝐴 = 𝐸1
That is, at least one of the relays must close for current to flow. Thus
P(A)=1-P(A)=1-P(E1)=1-0.04=0.96 which is the same as
P(E2)+P(E3)+P(E4)
1. If A and B are two independent events, P(A) = 0.6 and P(B) = 0.7,
ത is equal to
then 𝑃(𝐴ҧ ∪ 𝐵)
(a) 0.88
(b) 0.58
(c) 0.7
(d) None of these
b