Compound Events (Addition Rule) 1

Years

Year 12

Probability

Compound Events (Addition Rule)

Compound Events (Addition Rule) 1

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Question 1 of 8

1. Question
A normal six-sided dice is rolled. Identify which of the following are mutually exclusive events:
- Even or Odd
- Even or `>``5`
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Events are mutually exclusive if they cannot occur simultaneously.

Rolling the dice and getting an Even or Odd number.

Compare the results for each separate event

Sample Space of rolling a dice: `1,2,3,4,5,6`

Even:     `2,4,6`

Odd:     `1,3,5`

None of the results occurred simultaneously or is repeated. Hence, these are mutually exclusive events.

Rolling the dice and getting an Even or `>``5` number of dots.

Compare the results for each separate event

Sample Space of rolling a dice: `1,2,3,4,5,6`

Even:     `2,4,``6`

`>``5`:     `6`

`6` dots occurred simultaneously and is repeated on the two events. Hence, these are NOT mutually exclusive events.

Getting an Even or Odd number of dots is a mutually exclusive event

Question 2 of 8

2. Question

A jar contains `9` Red, `4` Green and `2` Black marbles. Find the probability of drawing a marble at random and getting:

`(a)` Red or Green

`(b)` Red or Black

Write fractions in the format “a/b”

`(a)` (13/15)

`(b)` (11/15)

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Addition Rule (Mutually Exclusive Events)

$$\mathsf{P(A\:or\:B)}=\mathsf{P(A)}+\mathsf{P(B)}$$

Events are mutually exclusive if they cannot
occur simultaneously.

Probability Formula

$$\mathsf{P(E)}=\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$

`(a)` Find the probability drawing a Red or Green marble.

Start by finding the probability of drawing a Red marble

favourable outcomes`=``9` (`9` Red)

total outcomes`=``15` (`9` Red, `4` Green, `2` Black)

$$ \mathsf{P(Red)} $$	`=`	$$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$	Probability Formula

	`=`	$$\frac{\color{#e65021}{9}}{\color{#007DDC}{15}}$$	Substitute values

Next, find the probability of drawing a Green marble

favourable outcomes`=``4` (`4` Green)

total outcomes`=``15` (`9` Red, `4` Green, `2` Black)

$$ \mathsf{P(Green)} $$	`=`	$$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$	Probability Formula

	`=`	$$\frac{\color{#e65021}{4}}{\color{#007DDC}{15}}$$	Substitute values

Finally, add the two probabilities

$$ \mathsf{P(Red\:or\:Green)} $$	`=`	$$\mathsf{P(Red)}+\mathsf{P(Green)}$$	Addition Rule

	`=`	$$\frac{9}{15}+\frac{4}{15}$$	Substitute values

	`=`	$$\frac{13}{15}$$

`(b)` Find the probability drawing a Red or Black marble.

From part `(a)`, we have solved the probability of drawing a Red marble

$$ \mathsf{P(Red)} $$

`=`

$$\frac{9}{15}$$

Next, find the probability of drawing a Black marble

favourable outcomes`=``2` (`2` Black)

total outcomes`=``15` (`9` Red, `4` Green, `2` Black)

$$ \mathsf{P(Black)} $$	`=`	$$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$	Probability Formula

	`=`	$$\frac{\color{#e65021}{2}}{\color{#007DDC}{15}}$$	Substitute values

Finally, add the two probabilities

$$ \mathsf{P(Red\:or\:Black)} $$	`=`	$$\mathsf{P(Red)}+\mathsf{P(Black)}$$	Addition Rule

	`=`	$$\frac{9}{15}+\frac{2}{15}$$	Substitute values

	`=`	$$\frac{11}{15}$$

`(a) 13/15`

`(b) 11/15`

Question 3 of 8

3. Question

Find the probability of drawing from a standard deck of cards and getting:

`(a)` Black `8` or Hearts

`(b)` King or Red

Write fractions in the format “a/b”

`(a)` (15/52)

`(b)` (7/13, 28/52)

Hint

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Excellent!

Incorrect

Addition Rule (Mutually Exclusive Events)

$$\mathsf{P(A\:or\:B)}=\mathsf{P(A)}+\mathsf{P(B)}$$

Events are mutually exclusive if they cannot
occur simultaneously.

Probability Formula

$$\mathsf{P(E)}=\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$

`(a)` Find the probability of drawing a Black `8` or Hearts.

Start by finding the probability of drawing a Black `8`

favourable outcomes`=``2` (`8` of Spades, `8` of Clubs)

total outcomes`=``52` (a standard deck has `52` cards)

$$ \mathsf{P(Black\:8)} $$	`=`	$$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$	Probability Formula

	`=`	$$\frac{\color{#e65021}{2}}{\color{#007DDC}{52}}$$	Substitute values

Next, find the probability of drawing a Hearts card

favourable outcomes`=``13` (`13` Hearts cards)

total outcomes`=``52` (a standard deck has `52` cards)

$$ \mathsf{P(Hearts)} $$	`=`	$$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$	Probability Formula

	`=`	$$\frac{\color{#e65021}{13}}{\color{#007DDC}{52}}$$	Substitute values

Finally, add the two probabilities

$$ \mathsf{P(Black\:8\:or\:Hearts)} $$	`=`	$$\mathsf{P(Black\:8)}+\mathsf{P(Hearts)}$$	Addition Rule

	`=`	$$\frac{2}{52}+\frac{13}{52}$$	Substitute values

	`=`	$$\frac{15}{52}$$

`(b)` Find the probability of drawing a King or Red card.

Start by finding the probability of drawing a King

favourable outcomes`=``4` (each of the suits have a King card)

total outcomes`=``52` (a standard deck has `52` cards)

$$ \mathsf{P(King)} $$	`=`	$$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$	Probability Formula

	`=`	$$\frac{\color{#e65021}{4}}{\color{#007DDC}{52}}$$	Substitute values

Next, find the probability of drawing a Red card

favourable outcomes`=``26` (`13` Hearts, `13` Diamonds)

total outcomes`=``52` (a standard deck has `52` cards)

$$ \mathsf{P(Red)} $$	`=`	$$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$	Probability Formula

	`=`	$$\frac{\color{#e65021}{26}}{\color{#007DDC}{52}}$$	Substitute values

Remember that the events need to be mutually exclusive which means the same card can’t be counted twice.

`2` of the cards are counted twice, which are King of Hearts and King of Diamonds

Hence, subtract `2/52` from the final probability.

Finally, solve for the final probability

$$ \mathsf{P(King\:or\:Red)} $$	`=`	$$\mathsf{P(King)}+\mathsf{P(Red)}-\frac{2}{52}$$	Addition Rule

	`=`	$$\frac{4}{52}+\frac{26}{52}-\frac{2}{52}$$	Substitute values

	`=`	$$\frac{28}{52}$$

	`=`	$$\frac{7}{13}$$

`(a) 15/52`

`(b) 7/13`

Question 4 of 8

4. Question

Find the probability of rolling `2` normal six-sided dice and getting:

`(a)` Sum of `10`

`(b)` Sum of `7`

Write fractions in the format “a/b”

`(a)` (1/12, 3/36)

`(b)` (1/6, 6/36)

Hint

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Correct

Well Done!

Incorrect

Probability Formula

$$\mathsf{P(E)}=\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$

`(a)` Find the probability of getting a Sum of `10`.

Set up a lattice showing all possible sums for the two dice

This means there are `36` possible outcomes

Now, count how many times we can have a sum of `10`

This means there are `3` times that we can have a sum of `10`

Finally, find the probability of getting a sum of `10`

favourable outcomes`=``3`

total outcomes`=``36`

$$ \mathsf{P(sum}\:10) $$	`=`	$$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$	Probability Formula

	`=`	$$\frac{\color{#e65021}{3}}{\color{#007DDC}{36}}$$	Substitute values

	`=`	$$\frac{1}{12}$$

`(b)` Find the probability of getting a Sum of `7`.

Use the lattice from part `(a)` and count how many times we can have a sum of `7`

This means there are `6` times that we can have a sum of `7`

Finally, find the probability of getting a sum of `7`

favourable outcomes`=``6`

total outcomes`=``36`

$$ \mathsf{P(sum}\:7) $$	`=`	$$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$	Probability Formula

	`=`	$$\frac{\color{#e65021}{6}}{\color{#007DDC}{36}}$$	Substitute values

	`=`	$$\frac{1}{6}$$

`(a) 1/12`

`(b) 1/6`

Question 5 of 8

5. Question

A normal six-sided dice is rolled and a fair coin is tossed. Find the probability of getting an Even number of dots and Heads.

Write fractions in the format “a/b”

(¼, 1/4, 3/12)

Hint

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Incorrect

Probability Formula

$$\mathsf{P(E)}=\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$

Set up a lattice showing all possible results for rolling a dice and tossing a coin

This means there are `12` possible outcomes

Now, count how many times we can have an Even number of dots and Heads

This means there are `3` times that we can have an Even number of dots and Heads

Finally, solve for the probability

favourable outcomes`=``3`

total outcomes`=``12`

$$ \mathsf{P(Even\:and\:Heads)} $$	`=`	$$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$	Probability Formula

	`=`	$$\frac{\color{#e65021}{3}}{\color{#007DDC}{12}}$$	Substitute values

	`=`	$$\frac{1}{4}$$

`1/4`

Question 6 of 8

6. Question

Find the probability of tossing `2` coins and getting:

`(a) 2` Tails

`(b)` Heads and Tails

Write fractions in the format “a/b”

`(a)` (1/4, ¼)

`(b)` (½, 1/2, 2/4)

Hint

Help Video

Correct

Well Done!

Incorrect

Probability Formula

$$\mathsf{P(E)}=\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$

`(a)` Find the probability of getting `2` Tails.

Set up a lattice showing all possible results for tossing `2` coins

This means there are `4` possible outcomes

Now, count how many times we can have `2` Tails

This means there is `1` time that we can have `2` Tails

Finally, find the probability of getting `2` Tails

favourable outcomes`=``1`

total outcomes`=``4`

$$ \mathsf{P(2\:Tails)} $$	`=`	$$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$	Probability Formula

	`=`	$$\frac{\color{#e65021}{1}}{\color{#007DDC}{4}}$$	Substitute values

`(b)` Find the probability of getting Heads and Tails.

Use the lattice from part `(a)` and count how many times we can have Heads and Tails

This means there are `2` times that we can have Heads and Tails

Finally, find the probability of getting Heads and Tails

favourable outcomes`=``2`

total outcomes`=``4`

$$ \mathsf{P(Heads\:and\:Tails)} $$	`=`	$$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$	Probability Formula

	`=`	$$\frac{\color{#e65021}{2}}{\color{#007DDC}{4}}$$	Substitute values

	`=`	$$\frac{1}{2}$$

`(a) 1/4`

`(b) 1/2`

Question 7 of 8

7. Question

Find the probability of rolling `2` normal six-sided dice and getting:

`(a)` Sum of `9` or `10`

`(b)` Sum of `6` or `12`

Write fractions in the format “a/b”

`(a)` (7/36)

`(b)` (1/6, 6/36)

Hint

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Correct

Correct!

Incorrect

Addition Rule (Mutually Exclusive Events)

$$\mathsf{P(A\:or\:B)}=\mathsf{P(A)}+\mathsf{P(B)}$$

Events are mutually exclusive if they cannot
occur simultaneously.

Probability Formula

$$\mathsf{P(E)}=\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$

`(a)` Find the probability of getting a Sum of `9` or `10`.

Set up a lattice showing all possible sums for the two dice

This means there are `36` possible outcomes

Now, count how many times we can have a sum of `9` or `10`

Having a sum of `9` appears `4` time

Having a sum of `10` appears `3` times

Finally, find the probability of getting a sum of `9` or `10` using Addition Rule

favourable outcomes`=``4+3=7`

total outcomes`=``36`

$$ \mathsf{P(sum\:of\:9\:or\:10)} $$	`=`	$$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$	Probability Formula

	`=`	$$\frac{\color{#e65021}{7}}{\color{#007DDC}{36}}$$	Substitute values

`(b)` Find the probability of getting a Sum of `6` or `12`.

Use the lattice from part `(a)` and count how many times we can have a sum of `6` or `12`

Having a sum of `6` appears `5` times

Having a sum of `12` appears `1` times

Finally, find the probability of getting a sum of `6` or `12` using Addition Rule

favourable outcomes`=``5+1=6`

total outcomes`=``36`

$$ \mathsf{P(sum\:of\:6\:or\:12)} $$	`=`	$$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$	Probability Formula

	`=`	$$\frac{\color{#e65021}{6}}{\color{#007DDC}{36}}$$	Substitute values

	`=`	$$\frac{1}{6}$$

`(a) 7/36`

`(b) 1/6`

Question 8 of 8

8. Question

Find the probability of rolling `2` normal six-sided dice and getting `1` or `5` dots on the dice

Write fractions in the format “a/b”

(5/9, 20/36)

Hint

Help Video

Correct

Nice Job!

Incorrect

Addition Rule (Mutually Exclusive Events)

$$\mathsf{P(A\:or\:B)}=\mathsf{P(A)}+\mathsf{P(B)}$$

Events are mutually exclusive if they cannot
occur simultaneously.

Probability Formula

$$\mathsf{P(E)}=\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$

Set up a lattice showing all possible results for rolling two dice

This means there are `36` possible outcomes

Now, count how many times `1` or `5` dots will appear

`1` dot appear `11` times

`5` dots appear `11` times

Remember that each roll must only be counted once.

`2` rolls are counted twice since they result to having both `1` AND `5` dots.

Hence, subtract `2/36` from the final probability.

Solve for the final probability using Addition Rule

favourable outcomes`=``11+11=22`

total outcomes`=``36`

$$ \mathsf{P(1\:or\:5)} $$	`=`	$$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}-\frac{2}{36}$$	Probability Formula

	`=`	$$\frac{\color{#e65021}{22}}{\color{#007DDC}{36}}-\frac{2}{36}$$	Substitute values

	`=`	$$\frac{20}{36}$$

	`=`	$$\frac{5}{9}$$

`5/9`

Quizzes

Simple Probability (Theoretical) 1
Simple Probability (Theoretical) 2
Simple Probability (Theoretical) 3
Simple Probability (Theoretical) 4
Complementary Probability
Compound Events (Addition Rule) 1
Compound Events (Addition Rule) 2
Venn Diagrams (Mutually Inclusive)
Independent Events 1
Independent Events 2
Dependent Events (Conditional Probability)
Probability Tree (Independent Events) 1
Probability Tree (Independent Events) 2
Probability Tree (Dependent Events)

Even:		`2,4,6`
Odd:		`1,3,5`

Even:		`2,4,``6`
`>``5`:		`6`

Compound Events (Addition Rule) 1

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Quiz summary

Information

1. Question

Hint

Great Work!

2. Question

Hint

Keep Going!

Addition Rule (Mutually Exclusive Events)

Probability Formula

3. Question

Hint

Excellent!

Addition Rule (Mutually Exclusive Events)

Probability Formula

4. Question

Hint

Well Done!

Probability Formula

5. Question

Hint

Keep Going!

Probability Formula

6. Question

Hint

Well Done!

Probability Formula

7. Question

Hint

Correct!

Addition Rule (Mutually Exclusive Events)

Probability Formula

8. Question

Hint

Nice Job!

Addition Rule (Mutually Exclusive Events)

Probability Formula

Quizzes