Compound Events 1

Topics

Pre - Algebra

Probability

Compound Events (Addition Rule)

Compound Events 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:
- 1. Even or $>$ $5$
- 2. Even or Odd
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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

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

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)

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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 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

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)

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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

$10$ .

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

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)

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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 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

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)

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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

$2$ 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

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)

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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

$9$ 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

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)

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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

$1$ 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

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

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

$5/9$

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Quiz summary

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1. Question

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Great Work!

2. Question

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Keep Going!

Addition Rule (Mutually Exclusive Events)

Probability Formula

3. Question

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Excellent!

Addition Rule (Mutually Exclusive Events)

Probability Formula

4. Question

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Well Done!

Probability Formula

5. Question

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Keep Going!

Probability Formula

6. Question

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Well Done!

Probability Formula

7. Question

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Correct!

Addition Rule (Mutually Exclusive Events)

Probability Formula

8. Question

Hint

Nice Job!

Addition Rule (Mutually Exclusive Events)

Probability Formula

Quizzes

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

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