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Probability Tree (Independent Events) 1Probability Tree (Independent Events) 1
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Question 1 of 5
1. Question
Find the probability of tossing a normal coin twice and getting:`(a) 2` Tails`(b)` Heads and Tails (in any order)Write fractions in the format “a/b”-
`(a)` (¼, 1/4)`(b)` (½, 1/2, 2/4)
Hint
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Well Done!
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Probability Formula
$$\mathsf{P(E)}=\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$Addition Rule
$$\mathsf{P(A\:or\:B)}=\mathsf{P(A)}+\mathsf{P(B)}$$Product Rule
$$\mathsf{P(A\:and\:B)}=\mathsf{P(A)}\times\mathsf{P(B)}$$`(a)` Find the probability of getting `2` Tails.First, set up a probability tree showing all possible outcomes of tossing a coin twice
Find the probability of all outcomes$$ \mathsf{P(Heads)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{2}}$$ `1` Heads out of `2` sides $$ \mathsf{P(Tails)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{2}}$$ `1` Tails out of `2` sides Mark each outcome with its corresponding probability
Finally, solve for the probability by picking the branch that leads to `2` Tails and multiplying the probabilities along it
`1/2``times``1/2` `=` `1/4` Product Rule Therefore, the probability of throwing Tails twice is `1/4``(b)` Find the probability of getting Heads and Tails.Use the probability tree from part `(a)` and pick the branches that lead to Heads and Tails then multiply the probabilities along it
First branch (HT):`1/2``times``1/2` `=` `1/4` Product Rule Second branch (TH):`1/2``times``1/2` `=` `1/4` Product Rule Finally, add the solved probability for each branch`1/4``+``1/4` `=` `2/4` `=` `1/2` Addition Rule Therefore, the probability of getting Heads and Tails is `1/2``(a) 1/4``(b) 1/2` -
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Question 2 of 5
2. Question
A spinner is spun twice. Find the probability of the spinner landing on Blue at least once
Write fractions in the format “a/b”- (5/9)
Hint
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Probability Formula
$$\mathsf{P(E)}=\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$Complementary
Probability$$\mathsf{P(\dot{E})}=1-\mathsf{P(E)}$$Addition Rule
$$\mathsf{P(A\:or\:B)}=\mathsf{P(A)}+\mathsf{P(B)}$$Product Rule
$$\mathsf{P(A\:and\:B)}=\mathsf{P(A)}\times\mathsf{P(B)}$$First, set up a probability tree showing all possible outcomes of the spinner landing on Blue at least once
A circle represents `360°`. From the diagram, we can see the probability of landing on Pink is `60/360`.We can also see Green represents half the circle`.: \text(Green)=180/360`To find the probability of landing on Blue, simply subtract the probability of landing on Green `color(forestgreen)((180))` and Pink `color(pink)((60))` from `360`.`\text(Blue)=360-(color(forestgreen)(180)+color(pink)(60))=120``:. \text(Blue)=120/360`Mark each outcome with its corresponding probability
Note in a probability tree event, the fractions when added must equal `1``1/6+1/2+1/3=1/1=1`Finally, pick the branches that lead to outcomes where the spinner lands on Blue at least once
`P(PB)+P(GB)+P(BP)+P(BG)+P(BB)``=` `(1/6xx1/3)+(1/2xx1/3)+(1/3xx1/6)+(1/3xx1/2)+(1/3xx1/3)` `=` `1/18+1/6+1/18+1/6+1/9` `=` `5/9` Therefore, the probability of the spinner landing on Blue at least once is `5/9``5/9` -
Question 3 of 5
3. Question
Find the probability of tossing a normal coin thrice and getting:`(a)` Heads, Tails, Tails (in that order)`(b) 2` Tails and Heads (in any order)Write fractions in the format “a/b”-
`(a)` (⅛, 1/8)`(b)` (⅜, 3/8)
Hint
Help VideoCorrect
Nice Job!
Incorrect
Probability Formula
$$\mathsf{P(E)}=\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$Addition Rule
$$\mathsf{P(A\:or\:B)}=\mathsf{P(A)}+\mathsf{P(B)}$$Product Rule
$$\mathsf{P(A\:and\:B)}=\mathsf{P(A)}\times\mathsf{P(B)}$$`(a)` Find the probability of getting Heads, Tails, Tails.First, set up a probability tree showing all possible outcomes of tossing a coin thrice
Find the probability of all outcomes$$ \mathsf{P(Heads)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{2}}$$ `1` Heads out of `2` sides $$ \mathsf{P(Tails)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{2}}$$ `1` Tails out of `2` sides Mark each outcome with its corresponding probability
Finally, solve for the probability by picking the branch that leads to Heads, Tails, Tails and multiplying the probabilities along it
`1/2``times``1/2``times``1/2` `=` `1/8` Product Rule Therefore, the probability of throwing Heads, Tails, Tails twice in that order is `1/8``(b)` Find the probability of getting `2` Tails and `1` Heads.Use the probability tree from part `(a)` and pick the branches that lead to `2` Tails and `1` Heads then multiply the probabilities along it
First branch (HTT):`1/2``times``1/2``times``1/2` `=` `1/8` Product Rule Second branch (THT):`1/2``times``1/2``times``1/2` `=` `1/8` Product Rule Third branch (TTH):`1/2``times``1/2``times``1/2` `=` `1/8` Product Rule Finally, add the solved probability for each branch`1/8``+``1/8``+``1/8` `=` `3/8` Addition Rule Therefore, the probability of getting `2` Tails and `1` Heads is `3/8``(a) 1/8``(b) 3/8` -
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Question 4 of 5
4. Question
Find the probability of tossing a normal coin thrice and getting:`(a) 3` of a kind`(b)` at least `2` HeadsWrite fractions in the format “a/b”-
`(a)` (¼, 1/4, 2/8)`(b)` (½, 1/2, 4/8)
Hint
Help VideoCorrect
Excellent!
Incorrect
Probability Formula
$$\mathsf{P(E)}=\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$Addition Rule
$$\mathsf{P(A\:or\:B)}=\mathsf{P(A)}+\mathsf{P(B)}$$Product Rule
$$\mathsf{P(A\:and\:B)}=\mathsf{P(A)}\times\mathsf{P(B)}$$`(a)` Find the probability of getting `3` of a kind.First, set up a probability tree showing all possible outcomes of tossing a coin thrice
Find the probability of all outcomes$$ \mathsf{P(Heads)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{2}}$$ `1` Heads out of `2` sides $$ \mathsf{P(Tails)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{2}}$$ `1` Tails out of `2` sides Mark each outcome with its corresponding probability
Pick the branches that lead to `3` of a kind then multiply the probabilities along it
First branch (HHH):`1/2``times``1/2``times``1/2` `=` `1/8` Product Rule Second branch (TTT):`1/2``times``1/2``times``1/2` `=` `1/8` Product Rule Finally, add the solved probability for each branch`1/8``+``1/8` `=` `2/8` `=` `1/4` Addition Rule Therefore, the probability of getting `3` of a kind is `1/4``(b)` Find the probability of getting at least `2` Heads.Use the probability tree from part `(a)` and pick the branches that lead to at least `2` Heads then multiply the probabilities along it
First branch (HHT):`1/2``times``1/2``times``1/2` `=` `1/8` Product Rule Second branch (HTH):`1/2``times``1/2``times``1/2` `=` `1/8` Product Rule Third branch (HHT):`1/2``times``1/2``times``1/2` `=` `1/8` Product Rule Fourth branch (HHH):`1/2``times``1/2``times``1/2` `=` `1/8` Product Rule Finally, add the solved probability for each branch`1/8``+``1/8``+``1/8``+``1/8` `=` `4/8` `=` `1/2` Addition Rule Therefore, the probability of getting at least `2` Heads is `1/2``(a) 1/4``(b) 1/2` -
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Question 5 of 5
5. Question
Find the probability of tossing a normal coin thrice and getting `3` Heads or Heads, Tails, Heads in that order.Write fractions in the format “a/b”- (¼, 1/4, 2/8)
Hint
Help VideoCorrect
Fantastic!
Incorrect
Probability Formula
$$\mathsf{P(E)}=\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$Addition Rule
$$\mathsf{P(A\:or\:B)}=\mathsf{P(A)}+\mathsf{P(B)}$$Product Rule
$$\mathsf{P(A\:and\:B)}=\mathsf{P(A)}\times\mathsf{P(B)}$$First, set up a probability tree showing all possible outcomes of tossing a coin thrice
Find the probability of all outcomes$$ \mathsf{P(Heads)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{2}}$$ `1` Heads out of `2` sides $$ \mathsf{P(Tails)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{2}}$$ `1` Tails out of `2` sides Mark each outcome with its corresponding probability
Pick the branches that lead to the preferred outcomes then multiply the probabilities along it
First branch (HHH):`1/2``times``1/2``times``1/2` `=` `1/8` Product Rule Second branch (HTH):`1/2``times``1/2``times``1/2` `=` `1/8` Product Rule Finally, add the solved probability for each branch`1/8``+``1/8` `=` `2/8` `=` `1/4` Addition Rule Therefore, the probability of getting `3` Heads and Heads, Tails, Heads is `1/4``1/4`
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)