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Venn Diagrams (Mutually Inclusive)Venn Diagrams (Mutually Inclusive)
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Question 1 of 6
1. Question
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 onlyWrite fractions in the format “a/b”-
`(a)` (13/30)`(b)` (1/6, 5/30)`(c)` (7/15, 14/30)
Hint
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Incorrect
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 outcomesNow, get the number of animals that can crawl`8+5` `=` `13` This means there are `13` animals that can crawlFinally, find the probability of selecting an animal that can crawlfavourable 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 flyThis means there are `5` animals that can crawl and flyFind the probabilityfavourable 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 flyThis means there are `14` animals that can only flyFind the probabilityfavourable 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` -
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Question 2 of 6
2. Question
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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Incorrect
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 Queenfavourable 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 cardfavourable 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 cardfavourable 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 DiagramStart with plotting the intersection of Queen and HeartsPlot the number of leftover Queen cards by subtracting the intersection from the total of Queen cardsPlot the number of leftover Heart cards by subtracting the intersection from the total of Heart cardsTo 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
3. Question
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)
Hint
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Excellent!
Incorrect
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 dogfavourable 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 catfavourable 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 catfavourable 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 DiagramStart with plotting the intersection of homes with Dogs and CatsPlot the number of leftover homes with Dogs by subtracting the intersection from the total of homes with DogsPlot the number of leftover homes with Cats by subtracting the intersection from the total of homes with CatsTo 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 catsSimply subtract the number of homes with either dogs or cats from the total number of homes surveyed`43/50` -
Question 4 of 6
4. Question
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)
Hint
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Incorrect
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 DiagramStart with plotting the intersection of students playing both Golf and TennisPlot 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
5. Question
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)
Hint
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Keep Going!
Incorrect
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 Hopfavourable 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 RnBfavourable 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 RnBfavourable 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 DiagramStart with plotting the intersection of Hip Hop and RnBPlot 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
6. Question
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)
Hint
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Correct!
Incorrect
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 Germanfavourable 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 Frenchfavourable 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 Frenchfavourable 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 DiagramStart with plotting the intersection of tourists that speak both German and FrenchPlot 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`
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)