Compound Events (Addition Rule) 2

Question 1 of 8

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)

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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 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

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)

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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 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

$12$ 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)

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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 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

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)

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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

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)

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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

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)

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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 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

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)

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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

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

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

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)

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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

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

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

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

$5/18$

Compound Events (Addition Rule) 2

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Quiz summary

Information

1. Question

Hint

Correct!

Addition Rule (Mutually Exclusive Events)

Probability Formula

2. Question

Hint

Fantastic!

Addition Rule (Mutually Exclusive Events)

Probability Formula

3. Question

Hint

Nice Job!

Addition Rule (Mutually Exclusive Events)

Probability Formula

4. Question

Hint

Great Work!

Probability Formula

5. Question

Hint

Excellent!

Probability Formula

6. Question

Hint

Excellent!

Addition Rule (Mutually Exclusive Events)

Probability Formula

7. Question

Hint

Fantastic!

Addition Rule (Mutually Exclusive Events)

Probability Formula

8. Question

Hint

Exceptional!

Addition Rule (Mutually Exclusive Events)

Probability Formula

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