Okay, let's solve this step-by-step: 1) Define the random variable: X = Number of trips of 5 days or more per year 2) Write the probability distribution: x P(x) 0 0.06 1 0.70 2 0.20 3 0.03 3) Calculate the mean using the formula: Mean = Σx * P(x) 0 * 0.06 + 1 * 0.70 + 2 * 0.20 + 3 * 0.03 = 0.84 So the mean number of trips per year is 0.84.

1. Elementary Statistics
Chapter 5:
Discrete Probability
Distribution
5.1 Probability
Distributions
1

2. Chapter 5: Discrete Probability Distribution
5.1 Probability Distributions
5.2 Binomial Probability Distributions
2
Objectives:
• Construct a probability distribution for a random variable.
• Find the mean, variance, standard deviation, and expected value for a discrete
random variable.
• Find the exact probability for X successes in n trials of a binomial experiment.
• Find the mean, variance, and standard deviation for the variable of a binomial
distribution.

3. Notation: P(A) = the probability of event A , 0 ≤ P(A) ≤ 1
The sum of the probabilities of all the outcomes in the sample space is 1: 𝑃𝑖 = 1
Impossible Set: If an event E cannot occur (i.e., the event contains no members in the sample space), its probability
is 0. P(A) = 0
Sure (Certain) Set: If an event E is certain, then the probability of E is 1. P(E) = 1
Complementary Events: 𝐴: consists of all outcomes that are not included in the outcomes of event A.
𝑃 𝐴 = 1 − 𝑃(𝐴)
The actual odds against event A: O 𝐴 =
𝑃 𝐴
𝑃(𝐴)
, (expressed as form of a:b or “a to b”); a and b are integers.
The actual odds in favor of event A: O 𝐴 =
𝑃 𝐴
𝑃 𝐴
(the reciprocal of the actual odds against the even) If the odds
against A are a:b, then the odds in favor of A are b:a.
The payoff odds against event A: Payoff odds against event A =
net profit
amount bet
net profit = (Payoff odds against event A)(amount bet)
Value: Give the exact fraction or decimal or round off final decimal results to 3 significant digits.
3
Recall: 4.1 Basic Concepts of Probability

4. Addition rule: A tool to find P(A or B), which is the probability that either event A occurs or event B occurs (or
they both occur) as the single outcome of a procedure. The word “OR” in the Addition rule is associated with the
addition of probabilities: P(A or B) = P(A) + P(B) − P(A and B)
Multiplication rule: A tool to find P(A and B), which is the probability that event A occurs and event B occurs.
The word “and” in the multiplication rule is associated with the multiplication of probabilities.
Compound Event: A compound event is any event combining two or more simple events.
Disjoint (or mutually exclusive): Events A and B are disjoint (or mutually exclusive) if they cannot occur at the
same time. (That is, disjoint events do not overlap.)
P(A and B) = P(event A occurs in a first trial and event B occurs in a second trial)
P(B | A) represents the probability of event B occurring after it is assumed that event A has already occurred.
Independent: Two events A and B are independent if the occurrence of one does not affect the probability of the
occurrence of the other. (Several events are independent if the occurrence of any does not affect the probabilities
of the occurrence of the others.) If A and B are not independent, they are said to be dependent.
P(A | B)= P(A) & P(B | A) = P(B)
Sampling with replacement: Selections are independent events: P(A ∩ B) = P(A)  P(B)
Sampling without replacement: Selections are dependent events.: P(A and B) = P(A)  P(B | A)
Recall 4.2 Addition Rule and Multiplication Rule( ) ( ) 1P A P A 
4

5. Complements: The Probability of “At Least One”
When finding the probability of some event occurring “at least once,” we should understand the following:
• “At least one” has the same meaning as “one or more.”
• The complement of getting “at least one” particular event is that you get no occurrences of that event.
P(at least one occurrence of event A) = 1 − P(no occurrences of event A)
A conditional probability of an event is a probability obtained with the additional information that some other
event has already occurred.
Notation
P(B | A) denotes the conditional probability of event B occurring, given that event A has already occurred.
It can be found by assuming that event A has occurred and then calculating the probability that event B will
occur.
Incorrect assumption of: P(B | A) = P(A | B) is called confusion of the inverse.
5
Recall 4.3 Complements and Conditional Probability, and Bayes’Theorem
( )
Prob of A Given B: ( ) , ( ) 0
( )
( )
Prob of B Given A: ( ) ; ( ) 0
( )
P A B
P A B P B
P B
P A B
P B A P A
P A

 

 
( t least 1) 1 ( )P A P None 

6. 1. Multiplication Counting Rule (The fundamental counting rule) For a
sequence of events in which the first event can occur n1 ways, the second
event can occur n2 ways, the third event can occur n3 ways, and so on, the
total number of outcomes is n1 · n2 · n3 . . . . (For a sequence of two events
in which the first event can occur m ways and the second event can occur n
ways, the events together can occur a total of m∙ n ways.)
2. Factorial: is the product of all the positive numbers from 1 to a number.
3. Permutation: is an arrangement of objects in a specific order. Order
matters. (The letter arrangements of abc, acb, bac, bca, cab, and cba are all
counted separately as six different permutations.) When n different items
are available and r of them are selected without replacement, the number of
different permutations (order counts) is given by
4. Permutations Rule (When Some Items Are Identical to Others) The
number of different permutations (order counts) when n items are available
and all n of them are selected without replacement, but some of the items
are identical to others, is found as follows: n1 are alike, n2 are alike, . .
. , and nk are alike.
5. Combination: is a grouping of objects. Order does not matter. When n
different items are available, but only r of them are selected without
replacement, the number of different combinations (order does not matter)
is found as follows:
6
Recall 4.4 Counting
    1 2 1
items
n n n n r
r
    
 
!
!
n r
n
P
n r


 
!
! !
n r
n
C
n r r

 !
n rP
r

( , ) n
r n rn r C C 
1 2
!
! ! !k
n
n n n 
  ! 1 2 3 2 1,0! 1n n n n      
TI Calculator:
Factorial
1. Enter the value of n
2. Press Math
3. Select PRB
4. Select ! & Enter
TI Calculator:
Permutation / Combination
1. Enter the value of n
2. Press Math
3. Select PRB
4. Select 𝒏𝑷 𝒓, or 𝒏𝑪 𝒓
5. Enter the value of r & Enter

7. Combining Descriptive Methods and Probabilities
Construct probability distributions by presenting possible outcomes along
with the relative frequencies.
7
5.1 Probability Distributions

8. Key Concept: Random variable & Probability Distribution.
A probability histogram is a graph that visually depicts a probability distribution.
Find its important parameters such as mean, standard deviation, and variance.
Determine whether outcomes are significant (significantly low or significantly high).
Random Variable: A random variable is a variable (typically represented by x) that has a single
numerical value, determined by chance, for each outcome of a procedure.
Probability Distribution: A probability distribution is a description that gives the probability for each
value of the random variable. It is often expressed in the format of a table, formula, or graph. A
discrete probability distribution consists of the values a random variable can assume and the
corresponding probabilities of the values.
Discrete Random Variable (DRV): A discrete random variable has a collection of values that is finite
or countable. (If there are infinitely many values, the number of values is countable if it is possible to
count them individually, such as the number of tosses of a coin before getting heads.)
Continuous Random Variable: A continuous random variable has infinitely many values, and the
collection of values is not countable. (That is, it is impossible to count the individual items because at
least some of them are on a continuous scale, such as body temperatures.) 8
5.1 Probability Distributions

9. Probability Distribution Requirements (3):
1. There is a numerical (not categorical) random variable (RV)
x, and its values are associated with corresponding
probabilities.
2. ∑P(x) = 1 where x assumes all possible values. (The sum of
all probabilities must be 1, but sums such as 0.999 or 1.001
are acceptable because they result from rounding errors.)
3. 0 ≤ P(x) ≤ 1 for every individual value of the random
variable x. (That is, each probability value must be between 0
and 1 inclusive.)
9
5.1 Probability Distributions

10. Construct a probability distribution for rolling a single die.
10
Probability Distributions Outcome x P(x)
1 1/6
2 1/6
3 1/6
4 1/6
5 1/6
6 1/6
Example 2
Represent graphically the probability distribution for the
sample space for tossing 3 coins.
# of Heads: x P(x)
0 1/8
1 3/8
2 3/8
3 1/8
Example 1

11. Let’s consider tossing two coins, with the following random variable:
x = number of heads when two coins are tossed
Is this a probability distribution?
x: Number of Heads When Two
Coins Are Tossed
P(x)
0 0.25
1 0.50
2 0.25 11
Example 3 Probability Distributions
1. The variable x is a numerical random variable, and its values are associated
with probabilities.
Yes, this is.
x is a Discrete Random Variable:
It has 3 possible values {0, 1, 2}
3 is a finite number.
This satisfies the requirement of being finite.
2. ∑P(x) = 0.25 + 0.50 + 0.25 = 1
3. Each value of P(x) is between 0 and 1.

12. A Probability Histogram is a graph of PD that displays the possible values of a
DRV on a horizontal axis and the probabilities of those values on the vertical
axis. A vertical bar represents the probability of each value and its height is
equal to the probability.
Probability Histogram for Number of
Heads When Two Coins Are Tossed
12
5.1 Probability Distributions
Probability Formula:
A probability distribution could also be in the form of a formula. Consider the formula
We find: P(0) = 0.25, P(1) = 0.50, and P(2) = 0.25.
The probabilities found using this formula are the same as those in the table.
1
( ) , 0,1, 2
2(2 )! !
P x x or
x x
 


13. Hiring managers were asked to identify the biggest mistakes that job applicants
make during an interview, and the table below is based on their responses. Does
the table below describe a probability distribution?
x P(x)
Inappropriate attire 0.50
Being late 0.44
Lack of eye contact 0.33
Checking phone or texting 0.30
Total 1.57
13
Example 4 Probability Distributions
Solution: The table violates:
1st requirement:
x is not a numerical random variable.
2nd requirement: ∑P(x) = 1
The sum of the probabilities = 1.57
The table does not describe a
probability distribution.

14. Parameters of a Probability Distribution
Remember that with a probability distribution, we have a description of a population
instead of a sample, so the values of the mean, standard deviation, and variance are
parameters, not statistics.
The mean, variance, and standard deviation of a discrete probability distribution can be
found with the following formulas:
Mean (Expected Value), Variance & Standard Deviation of D.R.V x:
Mean (the expected value) of occurrence per repetition (it represents the average
value of the outcome):
14
5.1 Probability Distributions μ = 𝐸 𝑥 = 𝑥𝑝(𝑥)
2 2 2
2 2 2
2
2 2 2
( ) ( )
( ) ( ) ( ) ,
( ) ( ) ( 2 ) ( )
( ) [2 ( ) ( )]
( ) [2 1] ( )
Pr
E x xp x
x p x x p x
x p x x x p x
x p x xp x p x
x x p
f
p x x
oo

  
  
 
   
 
   
   
  
     

 
 
  
 
Mean: ( )x P x  
2 2 2
Variance: ( )x P x     

15. Example 5
Given a probability distribution for rolling a single die, find
a. the mean.
b. the variance and standard deviation.
15
Probability Distributions
( )x p x  
1 1 1 1 1 1
6 6 6 6 6 61 2 3 4 5 6           
21
3.5
6
 
Outcome x P(x)
1 1/6
2 1/6
3 1/6
4 1/6
5 1/6
6 1/6
2 2 2
( )x p x     
 
2 2 2 21 1 1 1
6 6 6 6
22 21 1
6 6
1 2 3 4
5 6 3.5
       
    
2
2.91667 
1.70783 
( )x p x  
2 2 2
( )x p x     
2
( ) ( )x p x    

16. Example 6
The probability distribution shows the number of trips of
5 days or more that American adults take per year. Find the
mean.
16
Probability Distributions
6% take no trips
70% take 1 trip
20% take 2 trips
3% take 3 trips
1% take 4 trips
     
   
0 0.06 1 0.70 2 0.20
3 0.03 4 0.01
  
 
1.23 days
( )x p x  
( )x p x  
2 2 2
( )x p x     
2
( ) ( )x p x    

17. Recall: The table describes the probability distribution for the number of heads when two coins
are tossed. A) Find the mean, variance, and standard deviation for the probability distribution
described. B) Use these results and the range rule of thumb to determine whether 2
heads is a significantly high number of heads.
17
Example 7 Probability Distributions
x: Number of Heads When Two
Coins Are Tossed
P(x)
0 0.25
1 0.50
2 0.25
Sum
( )x p x  
xP(x)  2
( )x P x    
0(.25) (0 − 1)20.25 = 0.25
1(.5) (1 − 1)20.50 = 0
2(.25) (2 − 1)2
0.25 = 0.25
1 0.50
( ) 1x p x   
2 2
( ) ( ) 0.5x p x      
0.5 0.707107  
Interpretation: If we were to collect
data on a large number of trials with
two coins tossed in each trial, we
expect to get a mean of 1.0 head. B) (µ ± 2σ) =
1 ± 2(0.7) =
−0.4 𝑡𝑜 2.4
2 2 2
( )x p x      2
( ) ( )x p x    
No, why?

18. Lottery tickets cost $2 and out of 10,000
tickets printed, 1000 have a prize of $5, 100
have a prize of $10, 5 have a prize of $1000,
and 1 has a prize of $5000.
a. Let X represents the Net Gain and write
the probability distribution for winning
the lottery.
b. Calculate the expected value and the
standard deviation.
18
Example 8
Numbers: Win
1000: $5
100: $10
5: $1000
1: $5000
8894: Losers
10,000 Tickets
Net Gain: x p(x) = # / Total
$3 1000 / 10000 = 0.1
$8 100 / 10000 = 0.01
$998 5 / 10000 = 0.0005
$4998 1 / 10000 = 0.0001
−$2 8894 / 10000 = 0.8894
xp(x) 𝒙 𝟐
𝒑(𝒙)
0.3 0.9
0.08 0.64
0.499 498.002
0.4998 2498.0004
−1.7788 3.5576
𝑥𝑝 𝑥 = −0.4
𝑥2
𝑝 𝑥 = 3001.1
So on average a player
(all players) loses $
0.4 per ticket with
standard deviation of
$54.78 .
2 2
( )x p x     
2
3001.1 ( 0.4) $54.78   
 
     
 
The amount of the prize The Cost
1000 100 5
5 10 1000
10000 10000 10000
1
5000 2 0.4
10000
E x  
 
  Alternatively:
( )x p x   2 2 2
( )x p x     
2
( ) ( )x p x    
( ) 0.4x p x    

19. Identifying Significant Results with Probabilities:
Significantly high number of successes:
x successes among n trials is a significantly high number of successes if the probability of x or more
successes is 0.05 or less.
x is a significantly high number of successes if P(x or more) ≤ 0.05
Significantly low number of successes:
x successes among n trials is a significantly low number of successes if the probability of x or fewer
successes is 0.05 or less.
x is a significantly low number of successes if P(x or fewer) ≤ 0.05
The value 0.05 is not absolutely rigid. Other values, such as 0.01, could be used too.
The Rare Event Rule for Inferential Statistics: If, under a given
assumption, the probability of a particular outcome is very small and the outcome occurs significantly
less than or significantly greater than what we expect with that assumption, we conclude that the
assumption is probably not correct.
19
Example: Observation: A coin is tossed 100 times and we get 98 heads in a row.
Assumption: The coin is fair. (a normal assumption when an ordinary coin is tossed)
If the assumption is true, we have observed a highly unlikely event.
According to The Rare Event Rule , therefore, what we just observed is evidence against the assumption.
On this basis, we conclude the assumption is wrong (the coin is loaded.)

20. You have $5 to place on a bet in a casino; let’s consider the
following two bets:
a. Roulette: Bet on the number 7 in roulette.
P(Losing) = 37/38, P(winning) = 1/38 , Net gain = $175
b. Craps: Bet on the “pass line” in the dice game of craps.
P(Losing) = 251/495, P(winning) = 244/495, Net gain = $5
Which bet is better? (results in a higher expected value)
20
Example 10
Event (a)
Roulette
x P(x) xp(x)
Lose
Win
Total
Event(b)
Craps
x P(x) xp(x)
Lose
Win
Total
Interpretation: The $5 bet: Roulette: E(x) = −26¢ & Craps: E(x) = −7¢.
The craps game is better in the long run, even though the roulette game provides
an opportunity for a larger payoff when playing the game once.
( )x p x   2 2 2
( )x p x      2
( ) ( )x p x    
−4.8684
4.6053
−0.26
−2.5354
−2.2646
−0.07
−5 37/38
175 1/38
−5 251/495
5 244/495
( )x p x  