Venn Diagrams (Mutually Inclusive)

Question 1 of 6

Jack, an insect collector, has put together a sample space of his insects by representing it as a Venn Diagram as shown below. Find the probability of choosing an insect at random and getting an insect that can:

$(a)$ Crawl

$(b)$ Crawl AND fly

$(c)$ Fly only

Write fractions in the format “a/b”

$(a)$ (13/30)

$(b)$ (1/6, 5/30)

$(c)$ (7/15, 14/30)

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Addition Rule (Non Mutually Exclusive Events)

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

Events are non mutually exclusive if they can
occur simultaneously.

Probability Formula

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

$(a)$ Find the probability of choosing an insect that can crawl.

First, add all values on the diagram to find the sample space

$3+8+5+14$

$=$

$30$

This means there are $30$ possible outcomes

Now, get the number of animals that can crawl

$8+5$

$=$

$13$

This means there are $13$ animals that can crawl

Finally, find the probability of selecting an animal that can crawl

favourable outcomes

$=$ $13$

total outcomes

$=$ $30$

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

	$=$	$\frac{\color{#e65021}{13}}{\color{#007DDC}{30}}$	Substitute values

$(b)$ Find the probability of choosing an insect that can crawl AND fly.

Get the number of animals that can crawl AND fly

This means there are $5$ animals that can crawl and fly

Find the probability

favourable outcomes

$=$ $5$

total outcomes

$=$ $30$ (as solved from part

$(a)$ )

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

	$=$	$\frac{\color{#e65021}{5}}{\color{#007DDC}{30}}$	Substitute values

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

$(c)$ Find the probability of choosing an insect that can only fly.

Get the number of animals that can only fly

This means there are $14$ animals that can only fly

Find the probability

favourable outcomes

$=$ $14$

total outcomes

$=$ $30$ (as solved from part

$(a)$ )

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

	$=$	$\frac{\color{#e65021}{14}}{\color{#007DDC}{30}}$	Substitute values

	$=$	$\frac{7}{15}$

$(a) 13/30$

$(b) 1/6$

$(c) 7/15$

Question 2 of 6

Find the probability of drawing from a standard deck of cards and getting either a Queen or a Heart card.

Write fractions in the format “a/b”

(4/13, 16/52)

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Addition Rule (Non Mutually Exclusive Events)

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

Events are non mutually exclusive if they can
occur simultaneously.

Probability Formula

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

First, find the probability of drawing a Queen

favourable outcomes

$=$ $4$ (each of the

$4$ suits have

$1$ Queen)

total outcomes

$=$ $52$ (a standard deck has

$52$ cards)

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

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

Next, find the probability of drawing a Heart card

favourable outcomes

$=$ $13$ (there are

$13$ Heart cards)

total outcomes

$=$ $52$ (a standard deck has

$52$ cards)

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

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

Now, find the probability of drawing a Queen AND Heart card

favourable outcomes

$=$ $1$ (there is

$1$ Queen of Hearts card)

total outcomes

$=$ $52$ (a standard deck has

$52$ cards)

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

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

Finally, substitute these probabilities into the Addition Rule

$\mathsf{P(Queen)}=\frac{4}{52}$

$\mathsf{P(Heart)}=\frac{13}{52}$

$\mathsf{P(Queen\:AND\:Heart)}=\frac{1}{52}$

$\mathsf{P(Queen\:OR\:Heart)}$	$=$	$\mathsf{P(Queen)+P(Heart)-P(Queen\:AND\:Heart)}$

	$=$	$4/52+13/52-1/52$

	$=$	$\frac{\color{#00880A}{16}}{52}$

	$=$	$4/13$

The probability of drawing either a Queen or a Heart card is

$4/13$

This problem can also be represented visually using a Venn Diagram

Start with plotting the intersection of Queen and Hearts

Plot the number of leftover Queen cards by subtracting the intersection from the total of Queen cards

Plot the number of leftover Heart cards by subtracting the intersection from the total of Heart cards

To check, add the numbers:

$3+1+12=$ $16$

This matches the final probability computed earlier which is

$\frac{\color{#00880A}{16}}{52}$

$4/13$ or

$16/52$

Question 3 of 6

A survey was taken in a territory of

$50$ homes about whether they have a dog or a cat. It was found that

$35$ homes have dogs,

$17$ homes have cats, and

$9$ homes have both cats and dogs. Find the probability of picking a home at random that has either a cat or a dog.

Write fractions in the format “a/b”

(43/50)

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Addition Rule (Non Mutually Exclusive Events)

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

Events are non mutually exclusive if they can
occur simultaneously.

Probability Formula

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

First, find the probability of picking a home that has a dog

favourable outcomes

$=$ $35$ (

$35$ homes have dogs)

total outcomes

$=$ $50$ (

$50$ homes were surveyed)

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

	$=$	$\frac{\color{#e65021}{35}}{\color{#007DDC}{50}}$	Substitute values

Next, find the probability of picking a home that has a cat

favourable outcomes

$=$ $17$ (

$17$ homes have cats)

total outcomes

$=$ $50$ (

$50$ homes were surveyed)

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

	$=$	$\frac{\color{#e65021}{17}}{\color{#007DDC}{50}}$	Substitute values

Now, find the probability of picking a home that has a dog AND a cat

favourable outcomes

$=$ $9$ (

$9$ homes have dogs and cats)

total outcomes

$=$ $50$ (

$50$ homes were surveyed)

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

	$=$	$\frac{\color{#e65021}{9}}{\color{#007DDC}{50}}$	Substitute values

Finally, substitute these probabilities into the Addition Rule

$\mathsf{P(Dogs)}=\frac{35}{50}$

$\mathsf{P(Cats)}=\frac{17}{50}$

$\mathsf{P(Dogs\:AND\:Cats)}=\frac{9}{50}$

$\mathsf{P(Dogs\:OR\:Cats)}$	$=$	$\mathsf{P(Dogs)+P(Cats)-P(Dogs\:AND\:Cats)}$	Addition Rule

	$=$	$35/50+17/50-9/50$	Substitute values

	$=$	$\frac{\color{#00880A}{43}}{50}$

The probability of picking a home that has either dogs or cats is

$43/50$

This problem can also be represented visually using a Venn Diagram

Start with plotting the intersection of homes with Dogs and Cats

Plot the number of leftover homes with Dogs by subtracting the intersection from the total of homes with Dogs

Plot the number of leftover homes with Cats by subtracting the intersection from the total of homes with Cats

To check, add the numbers:

$26+9+8=$ $43$

This matches the final probability computed earlier which is

$\frac{\color{#00880A}{43}}{50}$

Additionally, take note of the homes that have neither dogs nor cats

Simply subtract the number of homes with either dogs or cats from the total number of homes surveyed

$43/50$

Question 4 of 6

At a school, all students must play either Golf or Tennis, or both. It is found that the probability of picking a student that plays Golf is

$0.53$ and the probability of picking a student that plays Tennis is

$0.72$ . What is the probability of selecting a student at random and finding out that he plays both Golf and Tennis?

Write your answer as a decimal

(0.25)

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Addition Rule (Non Mutually Exclusive Events)

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

Events are non mutually exclusive if they can
occur simultaneously.

Probability Formula

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

First, label the given numbers

$\mathsf{P(Golf)}=0.53$

$\mathsf{P(Tennis)}=0.72$

$\mathsf{P(Golf\:or\:Tennis)}=1$ (all students MUST play either Golf or Tennis)

Substitute these probabilities into the Addition Rule and then compute for the missing probability

$\mathsf{P(Golf\:or\:Tennis)}$	$=$	$\mathsf{P(Golf)+P(Tennis)-P(Golf\:and\:Tennis)}$
$1$	$=$	$0.53+0.72-\mathsf{P(Golf\:and\:Tennis)}$
$1$	$=$	$1.25-\mathsf{P(Golf\:and\:Tennis)}$
$\mathsf{P(Golf\:and\:Tennis)}+1$	$=$	$1.25$
$\mathsf{P(Golf\:and\:Tennis)}$	$=$	$1.25-1$
$\mathsf{P(Golf\:and\:Tennis)}$	$=$	$0.25$

The probability of picking a student that plays both Golf and Tennis is

$0.25$

This problem can also be represented visually using a Venn Diagram

Start with plotting the intersection of students playing both Golf and Tennis

Plot the number of leftover students that play Golf by subtracting the intersection from the total of students that play Golf.

$(0.53-0.25=0.28)$

Plot the number of leftover students that play Tennis by subtracting the intersection from the total of students that play Tennis.

$(0.72-0.25=0.47)$

$0.25$

Question 5 of 6

A group of

$21$ musicians play either Hip Hop or RnB. Specifically,

$14$ prefer to play Hip Hop and

$17$ prefer to play RnB. What is the probability of picking a musician that prefers playing both Hip Hop and RnB?

Write fractions in the format “a/b”

(10/21)

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Addition Rule (Non Mutually Exclusive Events)

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

Events are non mutually exclusive if they can
occur simultaneously.

Probability Formula

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

First, find the probability of picking a musician that prefers to play Hip Hop

favourable outcomes

$=$ $14$ (

$14$ musicians prefer Hip Hop)

total outcomes

$=$ $21$ (

$21$ total musicians)

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

	$=$	$\frac{\color{#e65021}{14}}{\color{#007DDC}{21}}$	Substitute values

Next, find the probability of picking a musician that prefers to play RnB

favourable outcomes

$=$ $17$ (

$17$ musicians prefer RnB)

total outcomes

$=$ $21$ (

$21$ total musicians)

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

	$=$	$\frac{\color{#e65021}{17}}{\color{#007DDC}{21}}$	Substitute values

Now, find the probability of picking a musician that prefers to play either Hip Hop or RnB

favourable outcomes

$=$ $21$ (all

$21$ musicians play either HipHop or RnB)

total outcomes

$=$ $21$ (

$21$ total musicians)

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

	$=$	$\frac{\color{#e65021}{21}}{\color{#007DDC}{21}}$	Substitute values

Finally, substitute these probabilities into the Addition Rule and compute for the missing value

$\mathsf{P(HipHop)}=\frac{14}{21}$

$\mathsf{P(RnB)}=\frac{17}{21}$

$\mathsf{P(HipHop\:or\:RnB)}=\frac{21}{21}$

$\mathsf{P(HipHop\:or\:RnB)}$	$=$	$\mathsf{P(HipHop)+P(RnB)-P(HipHop\:and\:RnB)}$

$21/21$	$=$	$\frac{14}{21}+\frac{17}{21}-\mathsf{P(HipHop\:and\:RnB)}$

$21/21$	$=$	$\frac{31}{21}-\mathsf{P(HipHop\:and\:RnB)}$

$\mathsf{P(HipHop\:and\:RnB)}+\frac{21}{21}$	$=$	$31/21$

$\mathsf{P(HipHop\:and\:RnB)}$	$=$	$31/21-21/21$

$\mathsf{P(HipHop\:and\:RnB)}$	$=$	$10/21$

The probability of picking a musician that prefers to play both Hip Hop and RnB is

$10/21$

This problem can also be represented visually using a Venn Diagram

Start with plotting the intersection of Hip Hop and RnB

Plot the number of leftover musicians that prefer Hip Hop by subtracting the intersection from the total of musicians that play Hip Hop.

$(14-10=4)$

Plot the number of leftover musicians that prefer RnB by subtracting the intersection from the total of musicians that play RnB.

$(17-10=7)$

$10/21$

Question 6 of 6

On a bus of

$30$ tourists,

$13$ speak German and

$23$ speak French. If

$7$ tourists speak both French and German, then what is the probability that a tourist chosen at random speaks neither German or French?

Write fractions in the format “a/b”

(1/30)

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Addition Rule (Non Mutually Exclusive Events)

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

Events are non mutually exclusive if they can
occur simultaneously.

Probability Formula

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

First, find the probability of picking a tourist that speaks German

favourable outcomes

$=$ $13$ (

$13$ tourists speak German)

total outcomes

$=$ $30$ (

$30$ total tourists)

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

	$=$	$\frac{\color{#e65021}{13}}{\color{#007DDC}{30}}$	Substitute values

Next, find the probability of picking a tourist that speaks French

favourable outcomes

$=$ $23$ (

$23$ tourists speak French)

total outcomes

$=$ $30$ (

$30$ total tourists)

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

	$=$	$\frac{\color{#e65021}{23}}{\color{#007DDC}{30}}$	Substitute values

Now, find the probability of picking a tourist that speaks both German and French

favourable outcomes

$=$ $7$ (

$7$ tourists speak both German French)

total outcomes

$=$ $30$ (

$30$ total tourists)

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

	$=$	$\frac{\color{#e65021}{7}}{\color{#007DDC}{30}}$	Substitute values

Substitute these probabilities into the Addition Rule and compute for the missing value

$\mathsf{P(German)}=\frac{13}{30}$

$\mathsf{P(French)}=\frac{23}{30}$

$\mathsf{P(German\:and\:French)}=\frac{7}{30}$

$\mathsf{P(German\:or\:French)}$	$=$	$\mathsf{P(German)+P(French)-P(German\:and\:French)}$

	$=$	$13/30+23/30-7/30$

	$=$	$29/30$

The probability of picking a tourist that speaks either German or French is

$29/30$

Finally, get the complement of the recently computed probability to get the probability of picking a tourist that speaks neither German or French

$\mathsf{P(neither\:German\:nor\:French)}$	$=$	$1-\mathsf{P(German\:or\:French)}$	Complementary Probability

	$=$	$1-29/30$	Substitute values

	$=$	$1/30$

The probability of picking a tourist that speaks neither German or French is

$1/30$

This problem can also be represented visually using a Venn Diagram

Start with plotting the intersection of tourists that speak both German and French

Plot the number of leftover tourists that speak German by subtracting the intersection from the total of tourists that speak German.

$(13-7=6)$

Plot the number of leftover tourists that speak French by subtracting the intersection from the total of tourists that speak French.

$(23-7=16)$

Lastly, plot the number of tourists that speak neither German or French.

$(6+7+16=29)$

Since the total number of tourists on the bus should equal

$30$

$(30-29=1)$ ,

$1$ person speaks neither German or French.

$1/30$

Venn Diagrams (Mutually Inclusive)

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Quiz summary

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1. Question

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Well Done!

Addition Rule (Non Mutually Exclusive Events)

Probability Formula

2. Question

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Nice Job!

Addition Rule (Non Mutually Exclusive Events)

Probability Formula

3. Question

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Excellent!

Addition Rule (Non Mutually Exclusive Events)

Probability Formula

4. Question

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Fantastic!

Addition Rule (Non Mutually Exclusive Events)

Probability Formula

5. Question

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Keep Going!

Addition Rule (Non Mutually Exclusive Events)

Probability Formula

6. Question

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Correct!

Addition Rule (Non Mutually Exclusive Events)

Probability Formula

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