Probability and Statistics part 1.pdf

Probability and
Statistics
FE Exam Prep Course
© 2018 Professional Publications, Inc.

PROBABILITY AND STATISTICS
• Combinations and Permutations
• Sets
• Probability
• Laws of Probability
• Measures of Central Tendency
• Measures of Dispersion
• Expected Values
• Probability Density Functions
• Probability Distribution Functions
• Probability Distributions
• Sums of Random Variables
• Hypothesis Testing
• Linear Regression
© 2018 Professional Publications, Inc. 2
Lesson Overview

You will learn
• how to calculate combinations and
permutations
• about the laws of probability
• about the mean, mode, median, and
other measures of central tendency
• about standard deviation, variance,
and other measures of dispersion
• about probability density and
distribution functions
• about binomial distribution and
normal distribution
• about the central limit theorem
• about unit normal and t-distribution
tables and how to use them
• about hypothesis testing
• about linear regression and the
method of least squares
Learning Objectives

probability
the measure of how likely a particular
event or outcome is to occur
statistics
the analysis and interpretation of
numerical data
Introduction

factorial, n!
• product of all positive integers from
1 to n
• for example,
6! = 6 × 5 × 4 × 3 × 2 × 1 = 720
• equal to the number of ways of
arranging n distinct objects in a
sequence
Combinations and Permutations

All 13 spades from a deck of playing cards
are shuffled and dealt face up in a row.
Most nearly, the number of different
sequences that can result is
(A) 40,000,000
(B) 500,000,000
(C) 6,000,000,000
(D) 90,000,000,000
Poll: Combinations and Permutations

All 13 spades from a deck of playing cards
are shuffled and dealt face up in a row.
Most nearly, the number of different
sequences that can result is
(A) 40,000,000
(B) 500,000,000
(C) 6,000,000,000
(D) 90,000,000,000
Solution
The number of ways to arrange 13
distinct items in a sequence is
13! = 6,227,020,800 (6,000,000,000)
The answer is (C).
Poll: Combinations and Permutations

permutation
• an order-conscious subset of elements
taken from a set
• order is significant—abc and bca are
different permutations
For example, if a set contains a, b, c, and
d, 24 permutations of three elements are
possible: abc, abd, acb, acd, adb, adc,
bac, bad, bca, bcd, bda, bdc, cab, cad,
cba, cbd, cda, cdb, dab, dac, dba, dbc,
dca, and dcb.
The number of permutations of r items
from a set of n items (with r ≤ n) is

combination
• a subset of elements taken from a set
• order is not significant—abc and bca
are the same combination
For example, if a set contains a, b, c, and
d, four combinations of three elements
are possible: abc, abd, acd, and bcd.
The combination abc could also be
written as acb, bac, bca, cab, or cba. All
six represent the same combination.
The number of combinations of r items
from a set of n items (with r ≤ n) is

A digital keypad containing the digits 0
through 9 is used to unlock a door. The
code is known to be five digits long and to
contain no repeated digits. The order in
which the digits are entered is significant,
so that 12345 and 54321 are two different
codes. Most nearly, how many possible
codes are there?
(A) 5000
(B) 8000
(C) 30,000
(D) 50,000
Example: Combinations and Permutations

A digital keypad containing the digits 0
through 9 is used to unlock a door. The
code is known to be five digits long and to
contain no repeated digits. The order in
which the digits are entered is significant,
so that 12345 and 54321 are two different
codes. Most nearly, how many possible
codes are there?
(A) 5000
(B) 8000
(C) 30,000
(D) 50,000
Solution
Because the order of the digits matters,
the total number of permutations must be
calculated.
The answer is (C).
     
, !/ ! 10!/ 10 5 !
30,240 (30,000)
P n r n n r
   


A pizzeria offers nine different toppings.
Most nearly, how many different
combinations of three toppings are
there?
(A) 20
(B) 30
(C) 80
(D) 500

A pizzeria offers nine different toppings.
Most nearly, how many different
combinations of three toppings are
there?
(A) 20
(B) 30
(C) 80
(D) 500
Solution
The answer is (C).
 
 
 
 
 
, !
9,3
! ! !
9!
3! 9 3 !
84 80
P n r n
C
r r n r
 





set
a collection of elements
element
a single item or outcome
For example,
• Let A be the set of red items on a
table: A = (apple, mug, pen)
• Let B be the set of edible items on the
same table: B = (apple, coffee, cookie,
muffin)
Mug is an element of set A, muffin is an
element of set B, and apple is an element
of both set A and set B.
Sets

outcome
a possible result of an experiment or trial
event
a set of outcomes that satisfy a particular
condition
For example, if the experiment is rolling a
single die, the set of all possible
outcomes is (1, 2, 3, 4, 5, 6).
• Let event A be rolling an even number.
• Let event B be rolling a number
greater than four.
1 and 3 are elements of neither event.
2 and 4 are elements of event A.
5 is an element of event B.
6 is an element of both events A and B.
Sets

sample space, S
• the set containing all possible items or
outcomes in the situation being studied
• also called universe, U.
When flipping a coin, the sample space is
(heads, tails).
When rolling a die, the sample space is
(1, 2, 3, 4, 5, 6).
When drawing two marbles one at a time
from an urn containing only black marbles
and white marbles, the sample space is
(black-black, black-white, white-black,
white-white).
Sets

and are sets containing elements from
the sample space, S.
• is the union of sets and .
• is the intersection of and .
• is the complement of set .
• indicates that is a subset of
(that is, every element in is also in ).
Sets

classical definition of probability
• based on reasoning
• if there are n equally likely outcomes
and of those outcomes belong to
event A,
For example, one card is drawn at
random from a standard deck of 52
playing cards. The chance of drawing a
face card (king, queen, or jack) is
Probability
 
12
face card 0.23
52
A
n
P
n
  

definition from relative frequency
• based on repeated experiments
• if there are n trials and of the
outcomes of those trials belong to
event A,
For example, an urn contains unknown
numbers of black marbles and white
marbles. One marble at a time is drawn
at random and then replaced. After 200
trials, white marbles have been drawn 64
times. The probability is estimated as
Larger numbers of trials will tend to give
more accurate estimates.
Probability
→
 
64
white 0.32
200
A
n
P
n
  

laws of probability
• The probability of any event E is
between zero and one.
• The probability of an event occurring
and the probability of it not occurring
add up to one.
• If event A is a subset of event B, the
probability of A is less than or equal to
the probability of B.
Laws of Probability
 
0 1
P E
 
   
not 1
P E P E
 
   
if , then
A B P A P B
 

independent events
If the success or failure of event has no
affect on the probability of event , then
the two events are independent.
For example, a die is rolled twice.
• Event is rolling a six on the first roll
of a die.
• Event is rolling a six on the second
roll.
Events and are independent.
Laws of Probability

dependent events
If the success or failure of event affects
the probability of event , then the two
events are dependent.
For example, an urn contains two black
marbles and two white marbles. Two
marbles are drawn, one after the other,
without replacement.
• Event is getting a black marble on
the first draw.
• Event is getting a black marble on
the second draw.
The probability of B is different
depending on whether is successful.
Events and are dependent.
Laws of Probability

conditional probability
P(A|B) is the conditional probability that
A will occur given that B has already
happened.
P(B|A) is the conditional probability that
B will occur given that A has already
happened.
Conditional probabilities are needed only
for calculations involving dependent
events. If events A and B are
independent, then P(A|B) = P(A) and
P(B|A) = P(B).
Laws of Probability

law of total probability
The probability of either event A or event B happening is
Laws of Probability

law of compound (joint) probability
The probability of both event A and event B happening is
If event A and event B are independent, then the equation reduces to
Laws of Probability
     
,
P A B P A P B


One bowl contains eight white balls and
two red balls. Another bowl contains four
yellow balls and six black balls. What is
the probability of getting a red ball from
the first bowl and a yellow ball from the
second bowl on one random draw from
each bowl?
(A) 0.08
(B) 0.2
(C) 0.4
(D) 0.8
Example: Laws of Probability

One bowl contains eight white balls and
two red balls. Another bowl contains four
yellow balls and six black balls. What is
the probability of getting a red ball from
the first bowl and a yellow ball from the
second bowl on one random draw from
each bowl?
(A) 0.08
(B) 0.2
(C) 0.4
(D) 0.8
Solution
is equal to because events
A and B are independent.
The answer is (A).
         
,
2 4
10 10
0.08
P A B P A P B A P A P B
 
  
   
  


A bowl contains eight white balls, two red
balls, four yellow balls, and six black balls.
What is the probability of getting a red
ball on the first draw and a yellow ball on
the second draw?
(A) 0.021
(B) 0.042
(C) 0.21
(D) 0.42

A bowl contains eight white balls, two red
balls, four yellow balls, and six black balls.
What is the probability of getting a red
ball on the first draw and a yellow ball on
the second draw?
(A) 0.021
(B) 0.042
(C) 0.21
(D) 0.42
Solution
There are 20 total balls and two are red,
so for the first draw, P(A) = 2/20.
If the first draw is successful, on the
second draw there are only 19 balls left
and four yellow balls, so P(B|A) = 4/19.
The answer is (A).
     
2 4
, |
20 19
0.021
P A B P A P B A
  
    
  


Bayes’ theorem
for two dependent events A and Bj, gives
the relationship between P(A|Bj) and
P(Bj|A)
• and are dependent events.
• B1, B2, B3, ..., Bn are the outcomes that
make up event B.
• Bj is a particular outcome in event B.
• The denominator sums to P(A).
Laws of Probability

In a study of a new test, 98% of those
known to have a certain illness test
positive, while just 4% of those known
not to have the illness test positive. The
illness affects 0.2% of the population.
Most nearly, the percentage of people
who test positive who have the illness is
(A) 4.7%
(B) 14%
(C) 54%
(D) 97%

In a study of a new test, 98% of those
known to have a certain illness test
positive, while just 4% of those known
not to have the illness test positive. The
illness affects 0.2% of the population.
Most nearly, the percentage of people
who test positive who have the illness is
(A) 4.7%
(B) 14%
(C) 54%
(D) 97%
Solution
The answer is (A).
 
   
   
 
   
       
  
     
 
1
|
|
|
ill pos|ill
ill|pos
pos|ill ill pos|not ill not ill
0.002 0.98
0.98 0.002 0.04 0.998
0.0468 4.7%
j j
j n
i i
i
P B P A B
P B A
P A B P B
P P
P
P P P P









• Combinations and Permutations
• Sets
• Probability
• Laws of Probability
• Measures of Central Tendency
• Measures of Dispersion
• Expected Values
• Probability Density Functions
• Probability Distribution Functions
• Probability Distributions
• Sums of Random Variables
• Hypothesis Testing
• Linear Regression
Lesson Overview

Probability and Statistics part 1.pdf

Recommended

Recommended

More Related Content

Similar to Probability and Statistics part 1.pdf

Similar to Probability and Statistics part 1.pdf (20)

More from Almolla Raed

More from Almolla Raed (8)

Recently uploaded

Recently uploaded (20)

Probability and Statistics part 1.pdf