AMDM Unit 2 Review

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Unit 2:Probability

Concept and
Notation of

Probability – L1

Georgia Standard and Learning Target

MAMDMD1. Students will determine probability and
expected value to inform everyday decision making.
❑ Determine conditional probabilities and probabilities of

compound events to make decisions in problem situations.

❑ Use probabilities to make and justify decisions about risks in

everyday life.

❑ Calculate expected value to analyze mathematical fairness,

payoff, and risk.

Dependent Probability

• One deck card/ One container
• Denominator will subtract to 1 for the

following withdrawal item.

• Numerator will be the same if the

following event is different color/card.

• Numerator will subtract to 1 if the

following same color/card

Two events are said to be dependent of
each other: the probabilityB of one event
occurs depends on the probability A

Dependent Rule

Probability: P(X and Y) = P(X) x P(Y)

• Total (denominator)?
• 1st Drawn = color/total
• 2nd Drawn = color/(total – 1)
• 3rd Drawn = color/(total – 2)

Review: Dependent Probability

Consider a container with 2 green, 3 blue and 4 purpleballs. Determine
probability of:
a)

1st purple/ 2nd blue drawn/ 3rd green drawn

b)

1st blue/ 2nd blue / 3rd green drawn

Total: 2 + 3 + 4 = 9

a) P (1st purple/2nd blue/3rd green) =

b) P (1st blue/2nd blue /3rd green) =

With a different colors, the denominator – 1
With a same colors, the numerator – 1 and,

the denominator - 1

Multiple Choice

Consider a container with 7 green, 8 blue , and 9 purple balls.  If two balls are drawn without replacement.  Determine the probability?

P (1st Gree/2nd Blue/3rd Purple)

• $$\left(\frac{8}{24}\right)\left(\frac{7}{23}\right)\left(\frac{6}{22}\right)$$  

• $$\left(\frac{8}{24}\right)\left(\frac{9}{23}\right)\left(\frac{10}{22}\right)$$  

• $$\left(\frac{7}{24}\right)\left(\frac{8}{23}\right)\left(\frac{9}{22}\right)$$  

• $$\left(\frac{8}{24}\right)\left(\frac{7}{24}\right)\left(\frac{6}{24}\right)$$  

Multiple Choice

Consider a container with 7 green, 8 blue , and 9 purple balls.  If two balls are drawn without replacement.  Determine the probability?

P (1st Blue /2nd Blue/3rd Blue)

• $$\left(\frac{8}{24}\right)\left(\frac{7}{23}\right)\left(\frac{6}{22}\right)$$  

• $$\left(\frac{8}{24}\right)\left(\frac{9}{23}\right)\left(\frac{10}{22}\right)$$  

• $$\left(\frac{7}{24}\right)\left(\frac{8}{23}\right)\left(\frac{9}{22}\right)$$  

• $$\left(\frac{8}{24}\right)\left(\frac{7}{24}\right)\left(\frac{6}{24}\right)$$  

Dependent Probability

One pack of 52 cards. Firstly, Pick a “King” card,
and then pick a “Queen” without the replacement
(on the same deck of cards)

•What is the probability of a “King” card? Is
the same as an independent probability?
•How many cards are left in the pack now?
Why the pack of cards are not 52?
•What is the probability of picking a “Queen”?
•What the probability of picking a “king” card
first, and then picking a “queen” card.

Dependent Probability of picking cards

Ex1: Probability of picking a “Jack” firstly, then secondly picking a
“Queen”, and lastly a “King” without any replacement on a deck of
cards?

52 cards
Draw three cards on one
standards deck of cards

Dependent Probability of picking cards

Ex2: Probability of picking a “Jack” firstly, and then secondly picking
another “Jack” , and thirdly picking a “Ace” without any replacement on
a deck of cards?

52 cards
Draw three cards on one
standards deck of cards

Dependent Probability of picking cards

Ex3: What the probability of picking a “7” firstly, then secondly picking
a “Jack”, thirdly picking another “7” and lastly picking another “Jack”
and without any replacement on a deck of cards?

52 cardsDraw three cards on one

standards deck of cards

Multiple Choice

Determine the probability of picking a "Ace" then another "King", and then a "Queen" and finally also "Jack" cards from without any placement on one standard deck of cards

• $$\left(\frac{4}{52}\right)\left(\frac{3}{51}\right)\left(\frac{2}{50}\right)\left(\frac{4}{49}\right)$$  

• $$\left(\frac{4}{52}\right)\left(\frac{4}{51}\right)\left(\frac{4}{50}\right)\left(\frac{4}{49}\right)$$  

• $$\left(\frac{4}{52}\right)\left(\frac{3}{52}\right)\left(\frac{4}{52}\right)\left(\frac{2}{52}\right)$$  

• $$\left(\frac{4}{52}\right)\left(\frac{4}{52}\right)\left(\frac{4}{52}\right)\left(\frac{4}{52}\right)$$  

Multiple Choice

Determine the probability of picking a "Ace" then another "Ace", and then a "Ace" again and finally also a "Queen" cards from without any placement on one standard deck of cards

• $$\left(\frac{4}{52}\right)\left(\frac{4}{52}\right)\left(\frac{4}{52}\right)\left(\frac{4}{52}\right)$$  

• $$\left(\frac{4}{52}\right)\left(\frac{4}{51}\right)\left(\frac{4}{50}\right)\left(\frac{4}{49}\right)$$  

• $$\left(\frac{4}{52}\right)\left(\frac{3}{52}\right)\left(\frac{2}{52}\right)\left(\frac{1}{52}\right)$$  

• $$\left(\frac{4}{52}\right)\left(\frac{3}{51}\right)\left(\frac{2}{50}\right)\left(\frac{4}{49}\right)$$  

Independent Probability

Two events are said to be independent of each other:
the probability of one event occurs doesn’t affects
the probability of the other event occurring.
Example: you rolled a die and flipped a coin.

Independent Rule

• Tossing a coin: P(coin) = 50% = 0.5 = 1/2
• Rolling a die: P(die) = 1/6
• Picking a card: P(card) = 4/52

Tossing a coin: P(coin) = 50% = 0.5 = 1/2
Rolling a die: P(die) = 1/6
Picking a card: P(card) = 4/52

Tossing two dice:
P(2 dice) =

Tossing two coin:
P(2 coins) =

Independent
Probability

Picking two card:
P(2 cards) =

Multiple Choice

Determine the probability of picking a "Ace" then another "King", and then a "Queen" and finally also "Jack" cards from four standard decks of cards

• $$\left(\frac{4}{52}\right)\left(\frac{3}{51}\right)\left(\frac{4}{50}\right)\left(\frac{2}{49}\right)$$  

• $$\left(\frac{4}{52}\right)\left(\frac{4}{51}\right)\left(\frac{4}{50}\right)\left(\frac{4}{49}\right)$$  

• $$\left(\frac{4}{52}\right)\left(\frac{3}{52}\right)\left(\frac{4}{52}\right)\left(\frac{2}{52}\right)$$  

• $$\left(\frac{4}{52}\right)\left(\frac{4}{52}\right)\left(\frac{4}{52}\right)\left(\frac{4}{52}\right)$$  

Multiple Choice

Determine the probability of picking a "Ace" then another "Ace", and then a "Ace" and finally also "Jack" cards from four standard decks of cards

• $$\left(\frac{4}{52}\right)\left(\frac{3}{51}\right)\left(\frac{2}{50}\right)\left(\frac{4}{49}\right)$$  

• $$\left(\frac{4}{52}\right)\left(\frac{4}{51}\right)\left(\frac{4}{50}\right)\left(\frac{4}{49}\right)$$  

• $$\left(\frac{4}{52}\right)\left(\frac{3}{52}\right)\left(\frac{2}{52}\right)\left(\frac{4}{52}\right)$$  

• $$\left(\frac{4}{52}\right)\left(\frac{4}{52}\right)\left(\frac{4}{52}\right)\left(\frac{4}{52}\right)$$  

Independent Probability – Dice

Roll three standard six-sided dice, what is the
probability that altogether land on a 4 or less?

Die 1

Die 2

P(die 2) = 4/6

P(die 1)= 4/6

P(three dice on 4 or less) = P(die 1) x P(die 2) x P(die 3)

independently

Die 3

P(die 3)= 4/6

•Less ⬄ count down
•Greater ⬄ count up

Ex9. Roll three standard six-sided dice, what is the
probability that altogether land on a 4 or less?

Independent Probability – Dice

Ex10. Roll three standard six-sided dice, what is the
probability that altogether land on a 4 or greater?

Die 1

Die 2

P(die 2) = 3/6

P(die 1)= 3/6

P(three dice on 4 or greater) = P(die 1) x P(die 2) x P(die 3)

P(three dice on 4 or less) =(4/6) x (4/6) x (4/6) = 64/216

independently

Die 3

P(die 3)= 3/6

Less
count
down

Greater
count

up

Multiple Choice

Determine probability of rolling four standard six-sided dice, what is the probability that all land on a 2 or greater

• $$\left(\frac{2}{6}\right)\left(\frac{2}{6}\right)\left(\frac{2}{6}\right)\left(\frac{2}{6}\right)$$  

• $$\left(\frac{3}{6}\right)\left(\frac{3}{6}\right)\left(\frac{3}{6}\right)\left(\frac{3}{6}\right)$$  

• $$\left(\frac{4}{6}\right)\left(\frac{4}{6}\right)\left(\frac{4}{6}\right)\left(\frac{4}{6}\right)$$  

• $$\left(\frac{5}{6}\right)\left(\frac{5}{6}\right)\left(\frac{5}{6}\right)\left(\frac{5}{6}\right)$$  

Multiple Choice

Determine probability of rolling four standard six-sided dice, what is the probability that all land on a 2 or less

• $$\left(\frac{2}{6}\right)\left(\frac{2}{6}\right)\left(\frac{2}{6}\right)\left(\frac{2}{6}\right)$$  

• $$\left(\frac{3}{6}\right)\left(\frac{3}{6}\right)\left(\frac{3}{6}\right)\left(\frac{3}{6}\right)$$  

• $$\left(\frac{4}{6}\right)\left(\frac{4}{6}\right)\left(\frac{4}{6}\right)\left(\frac{4}{6}\right)$$  

• $$\left(\frac{5}{6}\right)\left(\frac{5}{6}\right)\left(\frac{5}{6}\right)\left(\frac{5}{6}\right)$$  

Construct a Venn Diagram

Ms. Stringer conducted a survey
of his 9th grader homeroom. 17
students are taking both Econ
and Business. 19 students are
only taking Econ. 18 students are
only taking Business. 16 students
are taking other courses.
• Construct Venn Diagram. Which

one of the followings is correct
for the given Venn picture.

19
17

18

16

Construct a Venn Diagram

Mr. Jones conducted a survey of
his 9th grader homeroom. 20
students are taking both Econ
and Business. 30 students are
taking Econ (subtract to 20). 15
students are only taking
Business. 25 students are
taking other courses.
•Construct Venn Diagram.
Which one of the followings is
correct for the given Venn
picture.

10
20

15

25

30 – 20 = 10

Multiple Choice

Mr. Besson conducted a survey of his 10th grader homeroom.  6 students are taking both Math and Spanish.  14 students are only taking Math.  12 students are only taking Spanish.  10 students are taking other courses. Determine "v" (determine the total U"

• 6 + 14 + 12 + 10

• 6 + (14 - 6) + 12 + 10

• 6 + 14 + (12 - 6) + 10

• 6 + (14 - 6) + (12 - 6) + 10

Multiple Choice

Mr. Besson conducted a survey of his 10th grader homeroom.  6 students are taking both Math and Spanish.  14 students are taking Math.  12 students are only taking Spanish.  10 students are taking other courses. Determine "v" (determine the total U"

• 6 + 14 + 12 + 10

• 6 + (14 - 6) + 12 + 10

• 6 + 14 + (12 - 6) + 10

• 6 + (14 - 6) + (12 - 6) + 10

Multiple Choice

Mr. Besson conducted a survey of his 10th grader homeroom.  6 students are taking both Math and Spanish.  14 students are only taking Math.  12 students arey taking Spanish.  10 students are taking other courses. Determine "v" (determine the total U"

• 6 + 14 + 12 + 10

• 6 + (14 - 6) + 12 + 10

• 6 + 14 + (12 - 6) + 10

• 6 + (14 - 6) + (12 - 6) + 10

Multiple Choice

Mr. Besson conducted a survey of his 10th grader homeroom.  6 students are taking both Math and Spanish.  14 students are taking Math.  12 students are taking Spanish.  10 students are taking other courses. Determine "v" (determine the total U"

• 6 + 14 + 12 + 10

• 6 + (14 - 6) + 12 + 10

• 6 + 14 + (12 - 6) + 10

• 6 + (14 - 6) + (12 - 6) + 10

U (universal) =

P(H) = ?
i.

0.48

ii.

0.10

iii.

0.67

iv.

0.29

P(S) = ?
i.

0.48

ii.

0.10

iii.

0.67

iv.

0.29

iv

i

ii

iii

P(H) = (2 + 4)/21

4 + 2 + 8 + 7 = 21

P(S) = (2 + 8)/21

0.285…

0.476…

0.095…

0.667…

Probability’s concept and Notations

(Region shading for each event)

Probability is the ratio of number of
favorable outcomes to the total number {U}.
(for example: probability of C: P(C) = C/U

P(A) = A/U

P(B) = B/U

(A)

(B)

Multiple Choice

P(A) = ?

• (5)/27

• (5+9)/27

• (5+7)/27

• (5+7+9)/27

Multiple Choice

P(B) =?

• (5)/27

• (5+9)/27

• (5+7)/27

• (5+7+9)/27

Multiple Choice

P(A $$\cup$$  B) =?

• (5)/27

• (5+9)/27

• (5+7)/27

• (5+7+9)/27

Multiple Choice

P(A $$\cap$$  B) =?

• (5)/27

• (5+9)/27

• (5+7)/27

• (5+7+9)/27

Example 1: Probability of Complement

•

P(A’) = A’/U

P(B’) = B’/U

U (universal) =

P(H’) = ?
i.

0.52

ii.

0.90

iii.

0.33

iv.

0.71

P(S’) = ?
i.

0.52

ii.

0.90

iii.

0.33

iv.

0.71

iv

i

ii

iii

4 + 2 + 8 + 7 = 21

P(H)’ = (7 + 8)/21

0.714…

P(S)’ = (4 +7)/21

0.523…

0.904…

0.333…

Multiple Choice

Determine the probability of History complement:

P(H') = ?

• P(H)' =

  $$\frac{4+5+9}{21}$$  

• P(H)' = $$\frac{4\ +5}{21}$$  

• P(H)' =

  $$\frac{4+9}{21}$$  

• P(H)' =

  $$\frac{4}{21}$$  

Multiple Choice

Determine the probability of Science complement:

P(S') = ?

• P(S)' =

  $$\frac{4+5+9}{21}$$  

• P(S)' =

  $$\frac{4\ +5}{21}$$  

• P(S)' =

  $$\frac{4+9}{21}$$  

• P(S)' =

  $$\frac{4}{21}$$  

Multiple Choice

Determine the probability of (History and Science) complement:

$$P\left(H\cap S\right)'$$   = ?

• P(H and S)' =

  $$\frac{4+5+9}{21}$$  

• P(H and S)' =

  $$\frac{4\ +5}{21}$$  

• P(H and S)' =

  $$\frac{4+9}{21}$$  

• P(H and S)' =

  $$\frac{4}{21}$$  

Multiple Choice

Determine the probability of (History or Science) complement:

$$P\left(H\cup S\right)'$$   = ?

• P(H or S)' =

  $$\frac{4+5+9}{21}$$  

• P(H or S)' =

  $$\frac{4\ +5}{21}$$  

• P(H or S)' =

  $$\frac{4+9}{21}$$  

• P(H or S)' =

  $$\frac{4}{21}$$  

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