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1. 1
Statistics, Sample Test (Exam Review) Solution
Module 2: Chapters 4 & 5 Review
Chapter 4 Probability
Chapter 5: Discrete Probability Distribution
Chapter 4 Probability
1. Definitions:
a. A simple event is an outcome or event that cannot be further broken down.
b. A sample space is a procedure that consists of all possible sample events.
c. If two events are mutually exclusive, the probability that both will occur is
( ) 0
P A B
 =
d. The probability of an event is always:
a. between 0 and 1 0 ≤ P(A) ≤1
e. The sum of probabilities of all final outcomes of an experiment is always
1
( ) 1
n
i
i
P x
=
=

2. Answer the following:
a. The number of Combinations of n items selected n at a time is
nCr =
𝑛!
(𝑛−𝑟)!𝑟!
nCn =
𝑛!
(𝑛−𝑛)!𝑛!
=
𝑛!
0!𝑛!
=
𝑛!
𝑛!
= 1
b. The number of Permutations of n items selected 0 at a time is
nPr =
𝑛!
(𝑛−𝑟)!
nPo =
𝑛!
(𝑛−𝑜)!
=
𝑛!
𝑛!
= 1
c. A pizza parlor offers 10 different toppings; how many four topping pizzas (different
toppings) are possible?
210 four topping pizzas are possible
Order is not important → combination is used
nCr =
𝑛!
(𝑛−𝑟)!𝑟!
10C4 =
10!
(10−4)!4!
=
10𝑥9𝑥8𝑥7
4𝑥3𝑥2
= 210

2. 2
d. How many 6-letter code words can be made from the 26 letters of the alphabet if no
letter can be used more than once in the code word?
165, 765, 600 6-letter code words can be made
Order is important, no letter can be used more than once → permutation is used
26P6 =
26!
(26−6)!
=
26𝑥25𝑥24𝑥23𝑥22𝑥21
1
= 165,765, 600
3. Answer the following:
a. A quiz consists of 3 true-false questions, how many possible answer keys are there?
Write out the sample space and tree diagram.
8 possible answer keys
23
= 8
Sample Space: {TTT, TFF, FTF, FFT, FTT, TFT, TTF, FFF}
b. The sample space for tossing 5 coins consists of how many outcomes? Write out the
sample space.
Each coin has 2 outcomes: 25
= 32
Since there are only 2 possible outcome for each coin, Tail (T) or Head
(H), and there are 5 coins, Number of outcomes will be = 2 x 2 x 2 x 2 x 2
= 25 = 32
In general: Size of Sample Space = (# of outcomes per stage) # of stages
Sample Space:
{HHHHH, TTTTT, HHHHT, HHHTH, HHTHH, HTHHH, THHHH,
TTTTH, TTTHT, TTHTT, THTTT, HTTTT, HHHTT, HHTTH, HTTHH,
TTHHH, HHTHT, HTHHT, THHHT, HTHTH, THTHH, THHTH, TTTHH,
TTHHT, THHTT, HHTTT, TTHTH, THTTH, HTTTH, THTHT, HTHTT,
HTTHT}
4. A random sample of 100 people was asked if they were for or against the tax increase on
rich people. Of 60 males 45 were in favor, of all females 22 were in favor. Write the

3. 3
contingency table and answer the following questions. If one person is selected at
random, find the probability that:
For tax increase on
rich people (T)
Against tax increase on
rich people (A)
Total
Male (M) 45 15 60
Female (F) 22 18 40
Total 67 33 100
a) This person favors the tax increase on rich people.
P(T) = n(T) / n(S)
P(T) = 67/100 (or 0.67)
b) This person is a female.
P(F) = n(F) / n(S)
P(F) = 40/100 = 2/5 (or 0.4)
c) This person opposes the tax increase on rich people given that the person is a female.
P(A|F) = P(A ∩ F) / P(F)
P (A|F) = n(A ∩ F) / n(F)
P(A |F)= 18/40 = 9/20 (or 0.45)
d) This person is a male given that he favors the tax increase on rich people.
P(M|T) = P(M ∩ T) / P(T)
P(M|T) = n(M ∩ T) / n(T)
P(M|T) = 45/67 (or 0.6716)
e) This person is a female and favors the tax increase on rich people.
P (F∩ T) = 22/100 = 11/50 (or 0.22)
f) This person opposes the tax increase on rich people or is a female.
P(A U F) = P(A) + P(F) - P(A∩F)
P(A U F) = n(A)/n(S) + n(F)/n(S) – n(A∩F)/n(S)
P(A U F) = 33/100 + 40/100 – 18/100 = 55/100 = 11/20 (or 0.55)
g) Are the events “females” and opposes the tax increase on rich people independent?
Explain.

4. 4
Events “females” and opposes the tax increase on rich people are dependent because
occurrence of one of the events does affect the other event.
Proof: P (A |F) = 18/40 ≠ P (A) = 33/100
h) Are they mutually exclusive? Explain.
They are not mutually exclusive because a person can be both female and oppose the
tax increase on rich people. Those two events are not disjoint.
Proof: P (A∩F) = 18/100 ≠ 0
5. Answer the following:
a. Find the probability of getting the outcome of “Tails and 2” when a coin is tossed and
a die is rolled.
Independent Events: P(A∩B) = P(A) x P(B):
𝑃(𝑇𝑎𝑖𝑙𝑠 𝑎𝑛𝑑 2) = 𝑃(𝑇𝑎𝑖𝑙𝑠) × 𝑃(2) =
1
2
×
1
6
=
1
12
= 0.0833
b. A classic counting problem is to determine the number of different ways that the
letters of "PERSONNEL" can be arranged. Find that number.
Answer: Permutations Rule (Distinguishable Permutations) (When Some
Items Are Identical to Others)
𝑛!
𝑛1!𝑛2!…𝑛𝑘!
Requirements:
There are n items available, and some items are identical to others.
We select all of the n items (without replacement).
We consider rearrangements of distinct items to be different sequences.
There are 9 letters: n=9 P: 1 E:2 R:1 S:1 O:1 N:2 L:1
9!
1! 2! 1! 1! 1! 2! 1!
= 90,720
6. A box consists of 14 red and 36 blue markers. If we select 3 different markers randomly,
a. What is the probability that they are all red? (With replacement)
n(total) = 14 + 36 = 50
P(R) = n(R)/n(S) = 14/50 = 7/25
𝑃(1 𝑟𝑒𝑑) =
𝑇𝑜𝑡𝑎𝑙 𝑟𝑒𝑑
𝑇𝑜𝑡𝑎𝑙 𝑚𝑎𝑟𝑘𝑒𝑟𝑠
=
14
14+36
=
14
50
= 0.28
P(3 red) = 0.283
= 0.021952
b. What is the probability that they are all red? (Without a replacement) Draw a tree
diagram and label each branch.

5. 5
P(all 3 red) =
14
50
×
13
49
×
12
48
=
2184
117600
=
13
700
= 0.0186
2nd Method: P (all 3 red)=
𝐶3
14
𝐶3
50
7. If the probability of winning the race is 5/12,
a) What is the probability of losing the race?
1 – probability of winning
P(Ā) = 1- P(A) = 1 −
5
12
=
7
12
b) What are odds against winning?
𝑂(Ā) =
P(Ā)
P(A)
=
7
12
5
12
⁄ =
7
5
𝑜𝑟 7: 5
c) If the payoff odd is listed as 6:1, how much profit do you make if you bet $10 and
you win?
Payoff odds against event A = (net profit): (amount bet)
Net Profit = (Payoff odds) × (amount bet)
(10 )( 6) = $60
Red:
13/49
Red:
14/50
Blue:
36/50
Red:
12/48
Blue:
34/48
Blue:
35/49
Blue:
36/49
Blue:
36/48
Red:
14/49
Red:
13/48
Red:
13/48
Red:
14/48
Blue:
35/48
Blue:
35/48

6. 6
8. When two different people are randomly selected (from those in your class), find the
indicated probability (assume birthdays occur on the same day of the week with equal
frequencies).
a. Probability that two people are born on the same day of the week.
No particular day is specified, the first person can be born on any day.
𝑝(2𝑛𝑑 𝑝𝑒𝑟𝑠𝑜𝑛 𝑏𝑜𝑟𝑛 𝑜𝑛 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑑𝑎𝑦) =
1
7
𝑝(𝑏𝑜𝑡ℎ 𝑝𝑒𝑟𝑠𝑜𝑛𝑠 𝑏𝑜𝑟𝑛 𝑜𝑛 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑑𝑎𝑦) =
7
7
(
1
7
) =
1
7
b. Probability that two people are both born on Monday.
𝑝(1𝑠𝑡 𝑝𝑒𝑟𝑠𝑜𝑛 𝑏𝑜𝑟𝑛 𝑜𝑛 𝑀𝑜𝑛) =
1
7
𝑝(2𝑛𝑑 𝑝𝑒𝑟𝑠𝑜𝑛 𝑏𝑜𝑟𝑛 𝑜𝑛 𝑀𝑜𝑛) =
1
7
𝑝(𝑏𝑜𝑡ℎ 𝑝𝑒𝑟𝑠𝑜𝑛𝑠 𝑏𝑜𝑟𝑛 𝑜𝑛 𝑀𝑜𝑛) =
1
7
(
1
7
) =
1
49
9. How many different auto license plates are possible if the plate has:
Multiplication Counting Rule (The fundamental counting rule):
a) 2 letters followed by 4 numbers?
262
× 104
b) 3 letters – no repeats, followed by 3 numbers - repetition allowed?
26 × 25 × 24 × 103
c) 4 letters – repetition allowed, followed by 2 numbers – no repeats?
264
× 10 × 9
d) 4 places – each character is either a letter or a number?
26 𝐿𝑒𝑡𝑡𝑒𝑟𝑠 + 10 𝐷𝑖𝑔𝑖𝑡𝑠 = 36 𝐶ℎ𝑎𝑟𝑎𝑐𝑡𝑒𝑟𝑠 →
364

7. 7
10. In a first-grade school class, there are 10 girls and 8 boys. In how many ways can:
a. The students finish first, second and third in a foot race? (Assume no ties)
Order is important → permutation is used
𝑃𝑟
𝑛
= 𝑃3
18
= 18 × 17 × 16
b. The girls finish first and second in a geography contest? (Assume no ties)
Order is important → permutation is used
𝑃
𝑟
𝑛
= 𝑃2
10
= 10 × 9
c. 3 boys be selected for lunch duty?
Order is not important → combination is used
𝐶𝑟
𝑛
= 𝐶3
8
=
8!
3! (5!)
= 56
d. 6 students be selected for a hockey team?
Order is not important → combination is used
𝐶𝑟
𝑛
= 𝐶6
18
=
18!
6! (12!)
= 18,564
e. 5 students be selected: 3 boys and 2 girls?
Order is not important → combination is used
𝐶3
8
× 𝐶2
10
=
8!
3! (5!)
∙
10!
2! (8!)
= 2520
f. 4 girls be selected for a field trip?
Order is not important → combination is used
𝐶𝑟
𝑛
= 𝐶4
10
=
10!
4! (6!)
= 210

8. 8
Statistics, Sample Test (Exam Review)
Module 2: Chapters 4 & 5 Review
Chapter 5: Discrete Probability Distribution
1. Does the table describe probability distribution? What is the random variable, what are its
possible values, and are its values numerical?
Number of Girls in 3 Births
Number of girls x P(x)
0 0.125
1 0.375
2 0.375
3 0.125
Yes, there are 3 criteria:
1) The values of the random value x (which is the number of girls in three births) are numerical:
0, 1, 2, 3.
2) Their sum: 0.125 + 0.125 + 0.375 + 0.375 =1
3) The values of the random value x are between 0 and 1.
2. In a game, you pay 60 cents to select a 4-digit number. If you win by selecting the correct
4-digit number, you collect $3,000.
a) How many different selections are possible?
n = number of digits = 10: 0, 1, 2…,9
10 possible numbers in 4 places (numbers can repeat)
104
= 10,000
b) What is the probability of winning?
P(w) = n(w) / n(S) = 1/10,000
c) If you win, what is your net profit?
net profit = money gained – investment (Cost) = 3000 – 0.60 = $2999.40
d) Write the Probability Distribution of Net Profit if you win.

9. 9
Event x P(x) x P(x)
Lose −0.60 9,999/10,000 (0.9999) −0.59994
Win +2999.40 1/10,000 (0.0001) 0.29994
Note: Last column is not part of the Probability Distribution; it is used to calculate the
expected values in the next question.
e) Find the expected value and interpret.
E = μ = ∑ (x P(x))
= −0.6 × 0.9999 + 2999.4 × 0.0001 = −0.3
Anyone who plays the game loses about 30 cents on the average.
3. A pharmaceutical company receives large shipments of aspirin tablets. The acceptance
sampling plan is to randomly select and test 24 tablets. The entire shipment is accepted if
at most 2 tablets do not meet the required specifications. If a particular shipment of
thousands of aspirin tablets actually has a 5.0% rate of defects, what is the probability
that this whole shipment will be accepted?
Answer:
𝑩𝒊𝒏𝒐𝒎𝒊𝒂𝒍 𝒑𝒓𝒐𝒃𝒂𝒃𝒊𝒍𝒊𝒕𝒚 𝑫𝒊𝒔𝒕𝒓𝒊𝒃𝒖𝒕𝒊𝒐𝒏: 𝒏 = 𝟐𝟒, 𝒑 = 𝟎. 𝟎𝟓
𝑃(𝑥) =
𝑛!
(𝑛 − 𝑥)! 𝑥!
× 𝑝𝑥
× 𝑞𝑛−𝑥
𝑝(𝑥) =
24!
(24 − 𝑥)! 𝑥!
× 0.05𝑥
× 0.9524−𝑥
Given: 𝑛 = 24, 𝑝 = 0.05 & 𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒 𝑞 = 1 − 𝑝 = 0.95
P (x ≤ 2) (x = 0, 1, 2) =
P (0) + P (1) + P (2) = 0.884
4. It is known that 70% of managers of all companies suffer from job related stress. What is
the probability that in a sample of 20 managers?
a) Exactly 8 suffer from job related stress.
𝑩𝒊𝒏𝒐𝒎𝒊𝒂𝒍 𝒑𝒓𝒐𝒃𝒂𝒃𝒊𝒍𝒊𝒕𝒚 𝑫𝒊𝒔𝒕𝒓𝒊𝒃𝒖𝒕𝒊𝒐𝒏: 𝒏 = 𝟐𝟎, 𝒑 = 𝟎. 𝟕
𝑃(𝑥) =
𝑛!
(𝑛 − 𝑥)! 𝑥!
× 𝑝𝑥
× 𝑞𝑛−𝑥

10. 10
𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒 𝑞 = 1 − 𝑝 = 0.3
𝑝(𝑥) =
20!
(20 − 𝑥)! 𝑥!
× 0.7𝑥
× 0.320−𝑥
𝑃(8) =
20!
12! 8!
× 0.78
× 0.112
= 0.0038593
b) At most 8 suffer from job related stress.
𝑝(𝑥 ≤ 8): (𝑥 = 0,1, . . ,8) =
𝑝(0) + 𝑝(1)+… 𝑝(8) = 0.0051382
c) At least 9 suffer from job related stress.
P(𝑥 ≥ 9) (𝑥 = 9, 0, … 20),
𝑃(𝑥 ≥ 9) = P(9) + P(10) +… P(20) = 1– 𝑃(𝑥 ≤ 8) = 1 – 0.00514= 0.99486
d) Find the expected value.
E = μ = 𝑛 × 𝑝
E = 20 × 0.7 = 14
e) Find the standard deviation.
σ2
= 𝑛𝑝𝑞
σ2
= 10 (0.7) (0.3) = 4.2
σ = √4.2
σ = 2.05
f) Would it be unusual to claim that at most 7 managers from this sample suffer from
job related stress?
Range rule of thumb: μ ± 2σ
usual minimum: µ − 2σ =14 − 2(2.05) = 9.9
usual maximum: µ + 2σ = 14 + 2(2.05) = 18.1
7 ∉ (9.9, 18.1) → 𝑈𝑛𝑢𝑠𝑢𝑎𝑙
5. Find the probability of a couple having at least one girl among 3 children. (Discuss and
show all steps in two different methods.)
Method 1:
P (at least 1) = 1 – P (None)
= 1 – (1/2)3
= 7/8, Notice: p(g) = p(b) = 1/2
Method 2:

11. 11
Binomial Distribution: n = 3, x ≥1, p = q = ½
x = 1, 2, 3
P(1) + P(2) + P(3) = 1 – P(0)
𝐵𝑖𝑛𝑜𝑚𝑖𝑎𝑙 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑃(𝑥) =
𝑛!
(𝑛 − 𝑥)! 𝑥!
× 𝑝𝑥
× 𝑞𝑛−𝑥
𝑝(𝑥) =
3!
(3 − 𝑥)! 𝑥!
× 0.5𝑥
× 0.53−𝑥
𝑃(0) =
3!
3! 0!
× 0.50
× 0.53
P (0) = 1 × 1 × (0.5)3
P (x ≥ 1) = 1 – 0.53
= 1 – 0.125 = 0.875 (or 7/8)
6. If an alarm clock has a 0.9 probability of working on any given morning.
a) What is the probability that it will not work?
𝐶𝑜𝑚𝑝𝑖𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝐸𝑣𝑒𝑛𝑡 𝑃(𝑤
̅) = 1 − 𝑝(𝑤)
1 – 0.9 = 0.1 (or 10 percent)
b) What is the probability that 2 such alarm clocks will not work?
𝑃(𝑤1) = 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑦 𝑜𝑓 𝑎𝑛 𝑎𝑙𝑎𝑟𝑚 𝑤𝑜𝑟𝑘𝑖𝑛𝑔
𝑃(𝑤1
̅
̅̅̅) = 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑎𝑙𝑎𝑟𝑚 𝑁𝑂𝑇 𝑤𝑜𝑟𝑘𝑖𝑛𝑔
𝑃(𝑤1
̅
̅̅
̅ ∩ 𝑤2
̅̅̅̅) = 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝐴𝑁𝐷 𝑠𝑒𝑐𝑜𝑛𝑑 𝑎𝑙𝑎𝑟𝑚 𝑁𝑂𝑇 𝑤𝑜𝑟𝑘𝑖𝑛𝑔
𝑃(𝑤1
̅
̅̅̅ ∩ 𝑤2
̅̅̅̅) = 𝑃(𝑤1
̅
̅̅̅)𝑃(𝑤2
̅̅̅̅)
0.1 × 0.1 = 0.01
There is 1 % probability that 2 such alarm clocks will not work.
c) What is the probability of being awakened if you have 2 such alarm clocks?
Method 1:
Probability of being awakened = 1 – probability of not being awakened
1 – 0.01 = 0.99

12. 12
the probability of being awakened if you have 2 such alarm clocks is 99%.
Method 2: Addition Method
P(A U B) = P(A) + P(B) – P(A∩B) = 0.9 + 0.9 – (0.9 × 0.9) = 0.99
Method 3: Binomial Distribution: 𝑛 = 2 & 𝑝 = 0.9
𝐵𝑖𝑛𝑜𝑚𝑖𝑎𝑙 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑃(𝑥) =
𝑛!
(𝑛 − 𝑥)! 𝑥!
× 𝑝𝑥
× 𝑞𝑛−𝑥
𝑝(𝑥) =
2!
(2 − 𝑥)! 𝑥!
× 0.9𝑥
× 0.12−𝑥
Find the probability of:
𝑝(𝑥 ≥ 1) = 1 − 𝑃(0)
1 − 𝑃(0) = 1 −
2!
(2−0)!0!
× 0.90
× 0.12−0
= 1 − 0.01 = 0.99
7. During an NFL Season there were 256 games played with 1307 touchdowns scored.
(Poisson distribution)
a. What was the mean number of touchdowns (TD) scored in each game during the
season? (Round the answer to the nearest 0.0001)
The mean number of TDs in each game is: Mean: 𝝁 =
Number of TDs
Number of games
𝜇 =
1307
256
= 5.105469 ≈ 5.1055
𝑃(𝑥) =
𝜇𝑥
∙ 𝑒−𝜇
𝑥!
=
5.1055𝑥
∙ 𝑒−5.1055
𝑥!
b. On Jan 10, 2010 the Green Bay Packers and Arizona Cardinals played a playoff
game in which there were 13 touchdowns scored. What is the probability that a
random game would have that many or more touchdowns?
𝑝(𝑥 ≥ 13) = 1 − 𝑃(𝑥 ≤ 12)
= 1 − 0.9975911924 ≈ 0.0024
c. Complete the chart at the right. The first column lists the number of touchdowns in a
game, this is filled in already. The second column is for the predicted probability
that a game chosen at random will have that many touchdowns scored, calculate
these values, round these values to the closest 0.0001. The third column is for

13. 13
your best prediction about the number of games during the season that had that
many touchdowns scored, round these values to the closest whole number.
𝑃(𝑥) =
𝜇𝑥
∙ 𝑒−𝜇
𝑥!
=
5.1055𝑥
∙ 𝑒−5.1055
𝑥!
𝑃(0) =
5.10550
∙ 𝑒−5.1055
0!
= 0.006063
# TD Probability
(0.0001)
Whole Number: Predicted # of Games:
= 𝑷𝒓𝒐𝒃 𝑪𝒐𝒍𝒖𝒎𝒏 × 𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒈𝒂𝒎𝒆𝒔 ( 𝒐𝒓 𝟐𝟓𝟔)
0 0.006063496 0.006063496(256) = 1.5522 ≈ 2
1 0.03096 0.03096(256) = 7.925 ≈ 8
2 0.079025 0.079025(256) = 20.23 ≈ 20,
3 Continue!
4
5
6
7
8
9