MAT3026 - Probability and Statistics
Today's topics are:
• Probability
• Properties of Probability
• Conditional Probability and Independence
Lecture 4 1
Definition. A probability is a numerically valued function that assigns a number
P(A) to every event A of a sample space S, such that the following axioms hold:
(1) P(A) ≥ 0
(2) P(S) = 1
(3) If 𝐴1 , 𝐴2 , ... is a sequence of mutually exclusive events ( i.e. 𝐴𝑖 ∩ 𝐴𝑗 = Ø for
any i and j ), then
Theorem . If an experiment can result in any one of N different equally likely
outcomes, and if exactly n of these outcomes correspond to event A , then the
probability of event A is
𝑛
𝑃 𝐴 =
𝑁
Lecture 4 2
Example. A coin is tossed twice. What is the probability that at least one head
occurs?
The sample space for this experiment is S = {HH, HT, TH, TT}.
A represents the event of at least one head occurring is A= {HH, HT, TH}, then
𝑛 3
𝑃 𝐴 = = .
𝑁 4
Lecture 4 3
Lecture 4 4
Example. John is going to graduate from an industrial engineering department in
a university by the end of the semester. After being interviewed at two
companies he likes, he assesses that his probability of getting an offer from
company A is 0.8, and the probability that he gets an offer from company B is 0.6.
If on the other hand, he believes that the probability that he will get offers from
both companies is 0.5, what is the probability that he will get at least one offer
from these two companies?
P(A) = 0.8
P(B) = 0.6 𝑃 𝐴∪𝐵 =𝑃 𝐴 +𝑃 𝐵 −𝑃 𝐴∩𝐵
𝑃 𝐴 ∩ 𝐵 = 0.5
= 0.8 + 0.6 − 0.5 = 0.9
𝑃 𝐴∪𝐵 =?
Lecture 4 5
Example. What is the probability of getting a total 7 or 11 when a pair of dice are
tossed?
(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)
(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)
(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)
S = (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)
(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)
(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)
A : event that 7 occurs = { (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}
B : event that 11 occurs = { (5, 6), (6, 5)}
6
P(A) = 36
2
P(B) = 36 6 2 2
= + −0=
36 36 9
𝑃 𝐴∩𝐵 =0
𝑃 𝐴∪𝐵 =? Lecture 4 6
HOMEWORK: A pair of dice is tossed. Find the probability of getting
(a) a total of 8
(b) at most a total of 5.
Lecture 4 7
P(A∪B) = P(A)+P(B)-P(A∩B)
0.85 = 0.35 + P(B) - 0.19
P(𝐴′ ∩B‘) = P[(A∪B)′ ] = 1-P(A∪B)
P(B) = 0.69
P(A∪B) = 1-P(𝐴′ ∩B‘)
P(A∪B) = 1-0.15=0.85
Lecture 4 8
Example. If 2 of the 10 employees are female and 8 male, what is the probability
that exactly one female gets selected among the three?
2 8
1 2 = 56 = 0.47
10 120
3
Example. A package of 6 light bulbs contain 2 defective bulbs. If 3 bulbs are
selected for use, find the probability that none is defective.
4
3 = 4 = 0.2
6 20
3
Lecture 4 9
Example: If 3 books are picked at random from a shelf containing 5 novels, 3
books of poems, and a dictionary, what is the probability that
(a) the dictionary is selected?
8 1
2 1 = 28 = 0.33
9 84
3
(b) 2 novels and 1 book of poems are selected?
5 3
2 1 = 30 = 0.36
9 84
3
Lecture 4 10
The probability of an event B occurring when it is known that some event A has
occurred is called a conditional probability and is denoted by P B A . The symbol
P B A is usually read "the probability of B, given A."
Definition. The conditional probability of B, given A, is defined by
𝑃(𝐴 ∩ 𝐵)
𝑃 𝐵𝐴 = , 𝑃(𝐴) ≠ 0.
𝑃(𝐴)
Lecture 4 11
Example. The probability that a regularly scheduled flight departs on time is
𝑃(𝐷) = 0.83 ; the probability that it arrives on time is 𝑃(𝐴) = 0.82; and the
probability that it departs and arrives on time is 𝑃(𝐷 ∩ 𝐴) = 0.78. Find the
probability that a plane
(a) arrives on time given that it departed on time.
𝑃(𝐴 ∩ 𝐷) 0.78
𝑃 𝐴𝐷 = = = 0.94
𝑃(𝐷) 0.83
(b) departed on time given that it has arrived on time.
𝑃(𝐴 ∩ 𝐷) 0.78
𝑃 𝐷𝐴 = = = 0.95
𝑃(𝐴) 0.82
Lecture 4 12
Definition. Two events A and B are said to be independent if
Or
Theorem. Two events A and B are independent if and only if
𝑃 𝐴 ∩ 𝐵 = 𝑃 𝐴 𝑃(𝐵)
Lecture 4 13
1 3 1
Exercise. Let A and B be two events with 𝑃 𝐴 = 3 , 𝑃 𝐵′ = 4 and 𝑃 𝐴 ∪ 𝐵 = 2.
Find
𝑎) 𝑃 𝐴 ∩ 𝐵
𝑃 𝐴 ∪ 𝐵 = 𝑃 𝐴 + 𝑃 𝐵 − 𝑃(𝐴 ∩ 𝐵)
1 1 3 1
= + 1− −𝑃 𝐴∩𝐵 𝑃 𝐴∩𝐵 =
2 3 4 12
𝑏) 𝑃 𝐴′ ∩ 𝐵′
1 1
𝑃 𝐴′ ∩ 𝐵′ = 𝑃 (𝐴 ∪ 𝐵)′ = 1 − 𝑃 𝐴 ∪ 𝐵 = 1 − 2 = 2
𝑐) 𝑃(𝐵|𝐴)
𝑃(𝐴∩𝐵) 1/12 1
𝑃 𝐵𝐴 = = =4
𝑃(𝐴) 1/3
𝑑) 𝑃(𝐴 ∩ 𝐵′ )
𝑃 𝐴 = 𝑃 𝐴 ∩ 𝐵′ + 𝑃(𝐴 ∩ 𝐵)
1 ′ + 1 1
= 𝑃 𝐴 ∩ 𝐵 𝑃 𝐴 ∩ 𝐵′ = 4
3 12
𝑐) 𝑃(𝐴|𝐵′ )
𝑃(𝐴∩𝐵 ′ ) 1/4 1
𝑃 𝐴 𝐵′ = 𝑃(𝐵 ′ )
= 3/4 = 3
𝑒) Determine whether 𝐴 and 𝐵 are independent or not.
𝑃 𝐴∩𝐵 =𝑃 𝐴 𝑃 𝐵
1 1 1
= ∗ 𝐴 and 𝐵 are independent
12 3 4
Lecture 4 14
Theorem. If, in an experiment, the events A and B can both occur, then
𝑃 𝐴∩𝐵 =𝑃 𝐴 𝑃 𝐵 𝐴 .
Example. One bag contains 4 white balls and 3 black balls, and second bag
contains 3 white and 5 black balls. One ball is drawn from the first bag and placed
unseen in the second bag. What is the probability that a ball now drawn from the
second bag is black?
Solution. Let 𝐵1 , 𝐵2 and 𝑊1 represent, respectively, the drawing of a black ball
from bag1, a black ball from bag 2 and a white ball from bag 1.
Lecture 4 15
Example. A coin is biased so that a head is twice as likely to occur as a tail. If the
coin is tossed 3 times, what is the probability of getting 2 tails and 1 head?
Solution. The sample space for the experiment consists of the 8 elements,
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}.
Let A be the event of getting 2 tails and 1 head in the 3 tosses of an unfair coin.
Then
A = {TTH, THT, HTT}.
Assigning probabilities of w and 2w for getting a tail and a head, respectively, we
have
𝑃 𝐻 +𝑃 𝑇 =1
1
2𝑤 + 𝑤 = 1 𝑤 = 3.
1 2
Therefore, we obtain 𝑃 𝑇 = 3 and 𝑃 𝐻 = 3.
Since the outcomes on each of the 3 tosses are independent, it follows that
Lecture 4 16
HOMEWORK: The probability that Ayşe will be alive in 20 years is 0.7 and the
probability that Ali will be alive in 20 years is 0.9. If we assume independence for
both, what is the probability that neither will be alive in 20 years?
Lecture 4 17
Example. A random sample of 200 adults are classified below by sex and their level of
education attained. If a person is picked at random from this group, find the probability that
Education Male Female
(a) the person is a male and has a secondary education;
Elementary 38 45 83
28
𝑃 𝑀𝑎𝑙𝑒 & 𝑆𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦 = = 0.14 Secondary 28 50 78
200
College 22 17 39
(b) the person is a male, given that the person has a
88 112 200
secondary education;
𝑃 𝑀𝑎𝑙𝑒 & 𝑆𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦 28/200
𝑃 𝑀𝑎𝑙𝑒 𝑆𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦) = = = 0.36
𝑃 𝑆𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦 78/200
(c) the person does not have a college degree, given that the person is a female.
𝑃 𝐶𝑜𝑙𝑙𝑒𝑔𝑒 ′ 𝐹𝑒𝑚𝑎𝑙𝑒) = 1 − 𝑃 𝐶𝑜𝑙𝑙𝑒𝑔𝑒 𝐹𝑒𝑚𝑎𝑙𝑒)
𝑃 𝐶𝑜𝑙𝑙𝑒𝑔𝑒 & 𝐹𝑒𝑚𝑎𝑙𝑒 17/200
=1− = 1− = 0.85
𝑃 𝐹𝑒𝑚𝑎𝑙𝑒 112/200
Lecture 4 18
Lecture 4 19
Example. A group consists of 60% of girl students and 40% of boy students. 20%
of boy students and 50% of girl students are Turkish. If a student is taken at
random from this group,
(a) what is the probability that this student is Turkish?
(b) If at random selected student from this group is Turkish, what is the
probability that this student is a girl?
Solution. Consider the following events:
G: student is a girl B: student is a boy C: student is Turkish
Lecture 4 20
Example. In a certain assembly plant, three machines, B1 , B2 and B3 make
30%,45% and 25%, respectively, of the products. It is known from past experience
that 2%, 3%, and 2% of the product made by each machine, respectively, are
defective.
a) Find the probability that a randomly selected product is defective?
P(A) = (0.3)(0.02) + (0.45)(0.03) + (0.25)(0.02)
= 0.006 + 0.0135 + 0.005 = 0.0245
b) If a product were chosen randomly and found to be defective, what is the
probability that it was made by machine B3 ?
Lecture 4 21
