Compound Events (Addition Rule) 2

Years

Year 11

Probability

Compound Events (Addition Rule)

Compound Events (Addition Rule) 2

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Question 1 of 8

1. Question

Find the probability of spinning the arrow and getting:

`(a)` Blue or Pink

`(b)` Pink or Yellow

Write fractions in the format “a/b”

`(a)` (5/9, 200/360)

`(b)` (⅔, 2/3, 240/360)

Hint

Help Video

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 the arrow landing on Blue of Pink.

Start by finding the probability of getting Blue

favourable outcomes`=``90` (Blue section measures `90°`)

total outcomes`=``360` (whole circle measures `360°`)

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

	`=`	$$\frac{\color{#e65021}{90}}{\color{#007DDC}{360}}$$	Substitute values

Next, find the probability of getting Pink

favourable outcomes`=``110` (Pink section measures `110°`)

total outcomes`=``360` (whole circle measures `360°`)

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

	`=`	$$\frac{\color{#e65021}{110}}{\color{#007DDC}{360}}$$	Substitute values

Finally, add the two probabilities

$$ \mathsf{P(Blue\:or\:Pink)} $$	`=`	$$\mathsf{P(Blue)}+\mathsf{P(Pink)}$$	Addition Rule

	`=`	$$\frac{90}{360}+\frac{110}{360}$$	Substitute values

	`=`	$$\frac{200}{360}$$

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

`(b)` Find the probability of the arrow landing on Pink or Yellow.

From part `(a)`, we have solved the probability of getting Pink

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

`=`

$$\frac{110}{360}$$

Next, find the probability of getting Yellow

favourable outcomes`=``130` (Yellow section measures `130°`)

total outcomes`=``360` (whole circle measures `360°`)

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

	`=`	$$\frac{\color{#e65021}{130}}{\color{#007DDC}{360}}$$	Substitute values

Finally, add the two probabilities

$$ \mathsf{P(Pink\:or\:Yellow)} $$	`=`	$$\mathsf{P(Pink)}+\mathsf{P(Yellow)}$$	Addition Rule

	`=`	$$\frac{110}{360}+\frac{130}{360}$$	Substitute values

	`=`	$$\frac{240}{360}$$

	`=`	$$\frac{2}{3}$$

`(a) 5/9`

`(b) 2/3`

Question 2 of 8

2. Question

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

`(a)` Hearts or an Ace of Spades

`(b)` Diamonds or `5`

Write fractions in the format “a/b”

`(a)` (7/26, 14/52)

`(b)` (4/13, 16/52)

Hint

Help Video

Correct

Fantastic!

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 Hearts or an Ace of Spades card.

Start by finding 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

Next, find the probability of drawing an Ace of Spades

favourable outcomes`=``1` (there is only `1` Ace of Spades)

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

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

	`=`	$$\frac{\color{#e65021}{1}}{\color{#007DDC}{52}}$$	Substitute values

Finally, solve for the final probability

$$ \mathsf{P(Hearts\:or\:Ace\:of\:Spades)} $$	`=`	$$\mathsf{P(Hearts)}+\mathsf{P(Ace\:of\:Spades)}$$	Addition Rule

	`=`	$$\frac{13}{52}+\frac{1}{52}$$	Substitute values

	`=`	$$\frac{14}{52}$$

	`=`	$$\frac{7}{26}$$

`(b)` Find the probability of drawing a Diamonds or `5` card.

Start by finding the probability of drawing a Diamonds card

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

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

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

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

Next, find the probability of drawing a `5`

favourable outcomes`=``4` (each of the suits have a `5` card)

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

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

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

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

`1` of the cards is counted twice, which is the `5` of Diamonds

Hence, subtract `1/52` from the final probability.

Finally, solve for the final probability

$$ \mathsf{P(Diamonds\:or\:5)} $$	`=`	$$\mathsf{P(Diamonds)}+\mathsf{P(5)}-\frac{1}{52}$$	Addition Rule

	`=`	$$\frac{13}{52}+\frac{4}{52}-\frac{1}{52}$$	Substitute values

	`=`	$$\frac{16}{52}$$

	`=`	$$\frac{4}{13}$$

`(a) 7/26`

`(b) 4/13`

Question 3 of 8

3. Question

`12` cards are placed inside a hat, each marked with numbers `1`-`12`. Find the probability of drawing a card from this hat at random and getting:

`(a)` Odd or Multiple of `6`

`(b)` Even or Multiple of `5`

Write fractions in the format “a/b”

`(a)` (⅔, 2/3, 8/12)

`(b)` (7/12)

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

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

`(a)` Find the probability of drawing a card with an Odd number or a Multiple of `6`.

Start by finding the probability of drawing a Odd card

favourable outcomes`=``6` (`1,3,5,7,9,11`)

total outcomes`=``12` (`1,2,3,4,5,6,7,8,9,10,11,12`)

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

	`=`	$$\frac{\color{#e65021}{6}}{\color{#007DDC}{12}}$$	Substitute values

Next, find the probability of drawing a Multiple of `6`

favourable outcomes`=``2` (`6,12`)

total outcomes`=``12` (`1,2,3,4,5,6,7,8,9,10,11,12`)

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

	`=`	$$\frac{\color{#e65021}{2}}{\color{#007DDC}{12}}$$	Substitute values

Finally, add the two probabilities

$$ \mathsf{P(Odd\:or\:Multiple\:of\:6)} $$	`=`	$$\mathsf{P(Odd)}+\mathsf{P(Multiple\:of\:6)}$$	Addition Rule

	`=`	$$\frac{6}{12}+\frac{2}{12}$$	Substitute values

	`=`	$$\frac{8}{12}$$

	`=`	$$\frac{2}{3}$$

`(b)` Find the probability of drawing a card with an Even number of a Multiple of `5`.

Start by finding the probability of drawing an Even card

favourable outcomes`=``6` (`2,4,6,8,10,12`)

total outcomes`=``12` (`1,2,3,4,5,6,7,8,9,10,11,12`)

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

	`=`	$$\frac{\color{#e65021}{6}}{\color{#007DDC}{12}}$$	Substitute values

Next, find the probability of drawing a Multiple of `5`

favourable outcomes`=``2` (`5,10`)

total outcomes`=``12` (`1,2,3,4,5,6,7,8,9,10,11,12`)

$$ \mathsf{P(Multiple\:of\:5)} $$	`=`	$$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$	Probability Formula

	`=`	$$\frac{\color{#e65021}{2}}{\color{#007DDC}{12}}$$	Substitute values

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

`1` of the cards is counted twice, which is `10` (both Even and a Multiple of `5`)

Hence, subtract `1/12` from the final probability.

Finally, solve for the final probability

$$ \mathsf{P(Even\:or\:Multiple\:of\:5)} $$	`=`	$$\mathsf{P(Even)}+\mathsf{P(Multiple\:of\:5)}-\frac{1}{12}$$	Addition Rule

	`=`	$$\frac{6}{12}+\frac{2}{12}-\frac{1}{12}$$	Substitute values

	`=`	$$\frac{7}{12}$$

`(a) 2/3`

`(b) 7/12`

Question 4 of 8

4. Question

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

`(a)` Sum of `11`

`(b)` Sum of `5`

Write fractions in the format “a/b”

`(a)` (1/18, 2/36)

`(b)` (1/9, 4/36)

Hint

Help Video

Correct

Great Work!

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 `11`.

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 `11`

This means there are `2` times that we can have a sum of `11`

Finally, find the probability of getting a sum of `11`

favourable outcomes`=``2`

total outcomes`=``36`

$$ \mathsf{P(sum}\:11) $$	`=`	$$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$	Probability Formula

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

	`=`	$$\frac{1}{18}$$

`(b)` Find the probability of getting a Sum of `5`.

Use the lattice from part `(a)` and count how many times we can have a sum of `5`

This means there are `4` times that we can have a sum of `5`

Finally, find the probability of getting a sum of `5`

favourable outcomes`=``4`

total outcomes`=``36`

$$ \mathsf{P(sum}\:5) $$	`=`	$$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$	Probability Formula

	`=`	$$\frac{\color{#e65021}{4}}{\color{#007DDC}{36}}$$	Substitute values

	`=`	$$\frac{1}{9}$$

`(a) 1/18`

`(b) 1/9`

Question 5 of 8

5. Question

A fair coin is tossed and a spinner is spun. Find the probability of getting Tails and Pink.

Write fractions in the format “a/b”

(1/10)

Hint

Help Video

Correct

Excellent!

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 tossing a coin and spinning the spinner

This means there are `10` possible outcomes

Now, count how many times we can have Tails and Pink

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

Finally, solve for the probability

favourable outcomes`=``1`

total outcomes`=``10`

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

	`=`	$$\frac{\color{#e65021}{1}}{\color{#007DDC}{10}}$$	Substitute values

`1/10`

Question 6 of 8

6. Question

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

`(a)` Sum of `2` or `3`

`(b)` Sum of `4` or `5`

Write fractions in the format “a/b”

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

`(b)` (7/36)

Hint

Help Video

Correct

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 getting a Sum of `2` or `3`.

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 `2` or `3`

Having a sum of `2` appears `1` time

Having a sum of `3` appears `2` times

Finally, find the probability of getting a sum of `2` or `3` using Addition Rule

favourable outcomes`=``1+2=3`

total outcomes`=``36`

$$ \mathsf{P(sum\:of\:2\:or\:3)} $$	`=`	$$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$	Probability Formula

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

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

`(b)` Find the probability of getting a Sum of `4` or `5`.

Use the lattice from part `(a)` and count how many times we can have a sum of `4` or `5`

Having a sum of `4` appears `3` times

Having a sum of `5` appears `4` times

Finally, find the probability of getting a sum of `4` or `5` using Addition Rule

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

total outcomes`=``36`

$$ \mathsf{P(sum\:of\:4\:or\:5)} $$	`=`	$$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$	Probability Formula

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

`(a) 1/12`

`(b) 7/36`

Question 7 of 8

7. Question

Find the probability of rolling `2` normal six-sided dice and getting `2` or `3` dots on the dice

Write fractions in the format “a/b”

(5/9, 20/36)

Hint

Help Video

Correct

Fantastic!

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 `2` or `3` dots will appear

`2` dots appear `11` times

`3` dots appear `11` times

Remember that each roll must only be counted once.

`2` rolls are counted twice since they result to having both `2` AND `3` 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`

$$ \mathsf{P(2\:or\:3)} $$	`=`	$$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}-\frac{2}{36}$$	Probability Formula

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

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

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

`5/9`

Question 8 of 8

8. Question

Find the probability of rolling `2` normal six-sided dice and getting a product less than `6` when multiplying the number of dots.

Write fractions in the format “a/b”

(5/18, 10/36)

Hint

Help Video

Correct

Exceptional!

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 products for the two dice

This means there are `36` possible outcomes

Now, count how many times we can have a product less than `6`

There are `10` times that the product is less than `6`

Solve for the probability using Addition Rule

favourable outcomes`=``10`

total outcomes`=``36`

$$ \mathsf{P(Product<6)} $$	`=`	$$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$	Probability Formula

	`=`	$$\frac{\color{#e65021}{10}}{\color{#007DDC}{36}}$$	Substitute values

	`=`	$$\frac{5}{18}$$

`5/18`

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)