Download engstat q2 - DLSU

Council of Student Organizations
De La Salle University– Manila
Rm. 402 Bro. Connon Hall, De La Salle University
2401Taft Avenue, Manila
ENGSTAT Quiz 2 Reviewer
Prepared by: Ma. Elizabeth Ann L. Uy
PROBABILITY
A probability model consists of a sample space, S and an assignment of probability, P. Sample
space, S is the set of all possible outcomes of the random phenomenon. Sets of outcomes are called
Events. Any assignment of probability must obey the rules that state the basic properties of probability.
Probability Rules:
Rule 1: The probability P(A) of any event A satisfies 0 ≤ 𝑃(𝐴) ≤ 1
Rule 2: If S is the sample space in a probability model, then 𝑃(𝑆) = 1
Rule 3: Two events A and B are disjoints if they have outcomes in common and so can never occur
together. If a and B are disjoint, 𝑃(𝐴 𝑜𝑟 𝐵) = 𝑃(𝐴) + 𝑃(𝐵). This is the Addition Rule for Disjoint
Events.
Rule 4: The complement of ay event A is the event that A does not occur, written as A’.The complement
rule states that 𝑃(𝐴′ ) = 1 − 𝑃(𝐴)
Rule 5: The events A and B are independent if knowing that one occurs does not change the probability
that the other occurs. If A and B are independent, 𝑃(𝐴 𝑎𝑛𝑑 𝐵) = 𝑃(𝐴)𝑃(𝐵). This is the
Multiplication Rule for Independent Events.
EXAMPLE 1: Cellphones and Accidents Probabilities
Day
Probability
Sunday
0.03
Monday
0.19
Tuesday Wednesday Thursday
0.18
0.23
0.19
Friday
0.16
Saturday
0.02
What is the probability that an accident occurs on a weekend?
Answer:
Rule 3 (Disjoints Sets)
𝑷(𝑾) = 𝑷(𝒔𝒂𝒕) + 𝑷(𝒔𝒖𝒏) = 𝟎. 𝟎𝟐 + 𝟎. 𝟎𝟑 = 𝟎. 𝟎𝟓
Find the probability that a prne-related accident occurs on a weekday.
Answer:
Rule 3: 𝑷(𝑾𝒆𝒆𝒌𝒅𝒂𝒚𝒔) = 𝑷(𝑴) + 𝑷(𝑻) + 𝑷(𝑾) + 𝒑(𝑯) + 𝑷(𝑭) = 𝟎. 𝟗𝟓
Or Rule 4:
𝑷(𝑾𝒆𝒆𝒌𝒅𝒂𝒚𝒔) = 𝟏 − 𝑷(𝑾𝒆𝒆𝒌𝒆𝒏𝒅𝒔) = 𝟎. 𝟗𝟓
EXAMPLE 2: Educational levels of young adults
Choose a young adult (age 25 to 34 years old) at random. The probability is 0.12 that person
chosen did not complete high school, 0.31 that the person has a high school diploma but no further
education, and 0.29 that the person has at least a bachelor’s degree. A) What must be the probability that
a randomly chosen young adult has some education beyond high school but does not have a bachelor’s
degree? B) What is the probability that a randomly chosen young adult hast at least a high school
education?
Solution:
P(did not complete high school) = 0.12
P(has high school diploma but no further education) = 0.31
P(has at least a bachelor’s degree) = 0.29
TOTAL = 0.72
A) P(has high school diploma but no bachelor’s degree) = 1 – 0.72 = 0.28
B) P(young adult has at least a high school education) = 1- 0.12 = 0.88
EXAMPLE 3: Government Officials
Suppose that we take a simple random sample without replacement of two officials from the five
officials. A) Find the probability that we obtain the president and mayor. B) Find the probability that the
congressman is included in the sample.
President (P)
Vice President (VP)
Senate President (SP)
Congressman (C)
Mayor (M)
Solution:
PVP
PSP
VPSP
PC
VPC
SPC
P (President and Mayor) = 1/10
P (Congressman) = 4/10
VPM
SPM
CM
Answer:
PM
DISCRETE PROBABILITY AND DISTRIBUTION




Random Variable – variable whose value is a numerical outcome of a random phenomenon; no
bias
Discrete Random Variable – x has a finite number of possible values
Probability of Distribution of x – lists the values and their probabilities
Probabilities must satisfy two requirements:
o Every possibility is a number between 1 and 1.
o 𝑃1 + 𝑃2 + 𝑃3 + ⋯ + 𝑃𝑘 = 1
EXAMPLE 4: Grade Distribution
DLSU posts the grade distribution for its courses online. Studies in one section of ENGSTAT
received 31% A’s, 40% B’s, 20% C’s, 4% D’s and 5% F’s. Choose an ENGSTAT student at random. To “choose
at random” means to give every student the same chance to be chosen. The student’s grade on a fourpoint scale (w/ A=4) is a random variable x. Determine the probability that the student got a B or better.
Random Variable x
Probability Distribution
A
0.31
B
0.4
C
0.2
D
0.04
F
0.05
𝑷(𝑩 ∪ 𝑨) = 𝑷(𝑩) + 𝑷(𝑨) = 𝟎. 𝟒 + 𝟎. 𝟑𝟏 = 𝟎. 𝟕𝟏 = 𝟕𝟏%
PERMUTATION
n
𝑛!
𝑃=
r (𝑛 − 𝑟)!
EXAMPLE 5:
From among 10 employees, three are to be selected for travel to three out-of-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 assignment be made?
Solution:
n = 10 (sample space)
r = 3 (A, B, C)
𝑃=
1010!
= 720
(10
3 − 3)!
Answer: 720 ways
CONDITIONAL PROBABILITY (A|B)


Event A given B
One will not occur without the other
𝑃(𝐴|𝐵) =
𝑃(𝐴𝐵)
𝑃(𝐵)
𝑝𝑟𝑜𝑣𝑖𝑑𝑒𝑑 𝑃(𝐵) > 0
EXAMPLE 6:
Suppose that out of 100 students completing an introductory statistics course, 20 were business
majors. Ten receive A’s in the course, and three of these students were business majors. Determine the
probability that the student received an A, given that she’s a business major.
Solution:
Let A=event that the student received an A
Let B=event that the student is a business major
𝑃(𝐴|𝐵) =
𝑃(𝐴𝐵)
𝟑
=
𝑃(𝐵)
𝟐𝟎