Unlocking the Addition Rule of Probability: A Deep Dive

Event	Favorable Outcomes	Probability
$A$	$1,$ $2$	$\frac{2}{6} = \frac{1}{3}$
$B$	$5,$ $6$	$\frac{2}{6} = \frac{1}{3}$

Event

Favorable Outcomes

Probability

A

1,

2

\frac{2}{6} = \frac{1}{3}

B

5,

6

\frac{2}{6} = \frac{1}{3}

	Likes Classics	Does Not Like Classics
Likes Romances	$8$	$11$
Does Not Like Romances	$20$	$11$

Likes Classics

Does Not Like Classics

Likes Romances

8

11

Does Not Like Romances

20

11

	Likes Classics	Does Not Like Classics	Total
Likes Romances	$8$	$11$	$19 8 + 11 =$
Does Not Like Romances	$20$	$11$	$31 20 + 11 =$
Total	$28 8 + 20 =$	$22 11 + 11 =$	$50$

Likes Classics

Does Not Like Classics

Total

Likes Romances

8

11

19 8 + 11 =

Does Not Like Romances

20

11

31 20 + 11 =

Total

28 8 + 20 =

22 11 + 11 =

50

Event	Outcome(s)	Probability
Even	$2,$ $4,$ $6$	$P (A) = \frac{3}{6} = \frac{1}{2}$
Prime	$2,$ $3,$ $5$	$P (B) = \frac{3}{6} = \frac{1}{2}$
Even and prime	$2$	$P (A and B) = \frac{1}{6}$

Event

Outcome(s)

Probability

Even

2,

4,

6

P (A) = \frac{3}{6} = \frac{1}{2}

Prime

2,

3,

5

P (B) = \frac{3}{6} = \frac{1}{2}

Even and prime

2

P (A and B) = \frac{1}{6}

Notation	Meaning
$P (A) = a + c$	The probability of $A$ happening is $a + c .$
$P (B) = b + c$	The probability of $B$ happening is $b + c .$
$P (A \cup B) = a + b + c$	The probability of $A$ happening or $B$ happening is $a + b + c .$

Notation

Meaning

P (A) = a + c

The probability of

A

happening is

a + c .

P (B) = b + c

The probability of

B

happening is

b + c .

P (A \cup B) = a + b + c

The probability of

A

happening or

B

happening is

a + b + c .

Event	Favorable Outcomes	Probability
$A$	$2 ♣,$ $4 ♣,$ $6 ♣,$ $8 ♣,$ $10 ♣,$ $2 ♠,$ $4 ♠,$ $6 ♠,$ $8 ♠,$ $10 ♠,$ $2 ♢,$ $4 ♢,$ $6 ♢,$ $8 ♢,$ $10 ♢,$ $2 ♡,$ $4 ♡,$ $6 ♡,$ $8 ♡,$ $10 ♡$	$\frac{2 0}{5 2}$
$B$	$A ♣,$ $2 ♣,$ $3 ♣,$ $4 ♣,$ $A ♠,$ $2 ♠,$ $3 ♠,$ $4 ♠,$ $A ♢,$ $2 ♢,$ $3 ♢,$ $4 ♢,$ $A ♡,$ $2 ♡,$ $3 ♡,$ $4 ♡$	$\frac{1 6}{5 2}$
$C$	$A ♡,$ $2 ♡,$ $3 ♡,$ $4 ♡,$ $5 ♡,$ $6 ♡,$ $7 ♡,$ $8 ♡,$ $9 ♡,$ $10 ♡,$ $J ♡,$ $Q ♡,$ $K ♡$	$\frac{1 3}{5 2}$
$A and B$	$2 ♣,$ $4 ♣,$ $2 ♠,$ $4 ♠,$ $2 ♢,$ $4 ♢,$ $2 ♡,$ $4 ♡$	$\frac{8}{5 2}$
$A and C$	$2 ♡,$ $4 ♡,$ $6 ♡,$ $8 ♡,$ $10 ♡$	$\frac{5}{5 2}$
$B and C$	$A ♡,$ $2 ♡,$ $3 ♡,$ $4 ♡$	$\frac{4}{5 2}$
$A and B and C$	$2 ♡,$ $4 ♡$	$\frac{2}{5 2}$

Event

Favorable Outcomes

Probability

A

2 ♣,

4 ♣,

6 ♣,

8 ♣,

10 ♣,

2 ♠,

4 ♠,

6 ♠,

8 ♠,

10 ♠,

2 ♢,

4 ♢,

6 ♢,

8 ♢,

10 ♢,

2 ♡,

4 ♡,

6 ♡,

8 ♡,

10 ♡

\frac{2 0}{5 2}

B

A ♣,

2 ♣,

3 ♣,

4 ♣,

A ♠,

2 ♠,

3 ♠,

4 ♠,

A ♢,

2 ♢,

3 ♢,

4 ♢,

A ♡,

2 ♡,

3 ♡,

4 ♡

\frac{1 6}{5 2}

C

A ♡,

2 ♡,

3 ♡,

4 ♡,

5 ♡,

6 ♡,

7 ♡,

8 ♡,

9 ♡,

10 ♡,

J ♡,

Q ♡,

K ♡

\frac{1 3}{5 2}

A and B

2 ♣,

4 ♣,

2 ♠,

4 ♠,

2 ♢,

4 ♢,

2 ♡,

4 ♡

\frac{8}{5 2}

A and C

2 ♡,

4 ♡,

6 ♡,

8 ♡,

10 ♡

\frac{5}{5 2}

B and C

A ♡,

2 ♡,

3 ♡,

4 ♡

\frac{4}{5 2}

A and B and C

2 ♡,

4 ♡

\frac{2}{5 2}

Given that $A$ and $B$ are disjoint events, find $P (A or B)$ in the following cases. Give an exact answer in its simplest form.

P (A) = 0.3, P (B) = 0.1

P (A) = 0.5, P (B) = 0.2

P (A) = \frac{1}{3}, P (B) = \frac{1}{4}

It is given that events A and B are disjoint. This means they are mutually exclusive events. To calculate the probability that either one of these events occur, we can use the Addition Rule of Probability. P(A or B) = P(A)+ P(B) - P(A ⋂ B) However, since the events are disjoint, P(A ⋂ B) = 0. Because of this, the equation simplifies to P(A or B) = P(A)+ P(B). Let's calculate this probability.

Therefore, the probability that either one of the given events occur is 0.4

Just as in Part A, we are required to find the probability that either A or B occurs. Since these events are disjoint, the Addition Rule of Probability states the following. P(A or B) = P(A)+ P(B) Let's calculate this probability.

Therefore, the probability that either one of the given events occur is 0.4.

As in previous parts, to find the required probability we can use the Addition Rule of Probability. P(A or B) = P(A)+ P(B) Let's calculate this probability.

Therefore, the probability that either one of the given events occur is 712.

Dylan is conducting an experiment to determine how plants grow when exposed to different light sources. He has grown $400$ plants in the conditions indicated in the table below.

Plants exposed to infrared light	Plants exposed to ultraviolet light	Plants exposed to both infrared and ultraviolet light
$150$	$150$	$50$

What is the probability that a randomly selected plant had received either infrared or ultraviolet light? Answer with a fraction in its simplest form.

To find the probability that a randomly selected plant had received either infrared or ultraviolet light, we can use the Addition Rule of Probability. P(Aor B)=P(A)+P(B)-P(AandB) Notice that the probabilities appearing on the right-hand side of the formula can be calculated from the given information about the experiment's conditions. P(IR)&=150/400= 3/8 [1em] P(UV)&=150/400= 3/8 [1em] P(IR and UV)&=50/400= 1/8 With this information, we can calculate the probability that a plant had received either infrared or ultraviolet light.

Emily has set up a dart board and attached colored balloons to the board.

a dart board with twelve balloons of different colors stuck on it

Heichi throws a dart and pops one of them. What is the probability that the balloon was blue or orange? Answer with a fraction in its simplest form.

To calculate the probability that the balloon was either blue or orange, we can use the Addition Rule of Probability. P(Aor B)=P(A)+P(B) -P(A and B) Because all balloons are a single color, the events hitting a blue balloon and hitting an orange balloon cannot happen at the same time. Therefore P(blue and orange)=0, and in this case the equation can be simplified. P(blue or orange) =P(blue)+ P(orange) From the diagram we see that there are 12 balloons on the board. Of these 12, 3 are orange and 3 are blue. With this information we can determine the probabilities we need. P(blue)&=3/12= 1/4 [1em] P(orange)&=3/12= 1/4 Now we can determine the probability of hitting either a blue or an orange balloon.

Therefore, the probability that the balloon was either blue or orange is 12.

When considering a standard deck of cards, which of the following statements are correct?

i ii iii iv P (face card or club) P (spade or a seven) P (diamond or heart) None of the above are correct = \frac{1 1}{2 6} = \frac{5}{1 3} = \frac{1}{2}

To determine which of the given statements are correct, we can start by calculating the probability of the described events. Since the events involved are compound events, we can find their probability by using the Addition Rule of Probability. P(Aor B)=P(A)+P(B) -P(A and B) Recall that if two events A and B are disjoint, then P(A and B)=0. We can use the following formulas, according to each case. Disjoint events P(Aor B)=P(A)+P(B) [1em] Not disjoint events P(Aor B)=P(A)+P(B)-P(A and B) We will analyze each of the given statements one at a time.

i P(face card or club)

In a standard deck of cards we have 52 cards. They are divided into four suits, with each suit having 13 cards. For example, the cards that are suit of clubs are shown below.

Notice that 3 of these cards are face cards.

Therefore, by considering all face cards for the 4 different suits we have a total of 3* 4= 12. It is important to mention that since 3 of the 12 face cards are clubs, picking a face card and picking a club are not disjoint events. Therefore, the Addition Rule of Probability requires using the following formula. P(Aor B)=P(A)+P(B)-P(A and B) Using the previous information, we can calculate the required probabilities. P(club)&=13/52 [0.8em] P(face card)&=12/52 [0.8em] P(face card and club)&=3/52 Now we can determine the probability of picking either a face card or a club.

Therefore, the first statement is correct.

ii P(spade or a seven)

As explained in the previous part, there are 13 cards of each suit. Considering all 4 suits, there are 4 sevens, one of them is a spade. Therefore, picking a spade and picking a seven are not disjoint events. We will proceed in the same way as we did when calculating the probability of event ii. P(spade)&=13/52 [1em] P(seven)&=4/52 [1em] P(spade and a seven)&=1/52 Now let's calculate the probability of picking either a spade or a seven.

Therefore, statement ii is not true.

iii P(diamond or heart)

Since there are no cards that are both diamond and heart, picking a diamond and picking a heart are disjoint events. In this case the Addition Rule of Probability reduces to the formula given below. P(Aor B)=P(A)+P(B) Recalling that there are 13 cards of each suit, we can calculate the probabilities that we will need. P(diamond)&=13/52 [1em] P(heart)&=13/52 Now we can calculate the probability of picking either a diamond or a heart.

Therefore, statement iii is true.

In the following Venn diagram, the outcomes for each event are represented using dots.

a venn diagram with three overlapping circles

Which of the following statements are true?

i ii iii iv A and B are disjoint events A and C are disjoint events B and C are disjoint events None of the above are true

Calculate

P (A or B) .

Answer with a fraction in its simplest form.

Calculate

P (B or C) .

Answer with a fraction in its simplest form.

Recall that two events are disjoint or mutually exclusive if they cannot happen at the same time. We can verify which events are not disjoint by looking at the Venn diagram. If two events have outcomes in an overlapping region, then they can happen at the same time.

We can see that A and C can happen at the same time, as they have two outcomes in their overlapping region. This is also true for the overlapping region of B and C. We also see that there are no common outcomes in the overlapping region between A and B. Therefore, A and B are the only pair of disjoint events, and thus i is the only true statement.

In Part A we found out that A and B are disjoint events. Therefore, we can calculate the probability that either A or B happens by using the Addition Rule of Probability. P(A or B)=P(A)+P(B) To calculate the different probabilities we need, we can divide the favorable outcomes by the total number of possible outcomes. The total number of possible outcomes is the number of dots in the diagram. There are 15 dots in total. Furthermore, there are 5 dots in A and 6 dots in B. P(A)&=5/15 [1em] P(B)&=6/15 Now we can calculate P(A or B).

As explained in previous parts, B and C are not disjoint events. Therefore, the Addition Rule of Probability requires using the following formula. P(B or C)=P(B)+P(C)-P(BandC) We already know that we have a total of 15 possible outcomes. Of these, 7 are in C, 6 are in B, and 2 are in the overlapping area between B and C. With this information we can determine the probabilities that we need. P(B)&= 6/15 [0.9em] P(C)&= 7/15 [0.9em] P(B andC)&= 2/15 Now we can determine the probability that either B or C occurs.

Addition Rule in the Uniform Probability Model

Catch-Up and Review

Calculating the Chances of Winning a Card Game

Finding the Probability of a Compound Event When Rolling a Die

Hint

Solution

A Survey, Probabilities, and a Decision

Hint

Solution

Addition Rule of Probability

Proof

Calculating the Probability of the Union of Two Events

Hint

Solution

Practice Calculating Probabilities of the Union of Two Events

Calculating the Probability of the Union of Three Overlapping Events

Answer

Hint

Solution

Probability of Winning a Card Game with Multiple Winning Conditions

i P(face card or club)

ii P(spade or a seven)

iii P(diamond or heart)

Addition Rule in the Uniform Probability Model

Recommended exercises

	9 Theory slides
	8 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions