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Probability Tree (Independent Events)

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Probability Tree (Independent Events) 1

Probability Tree (Independent Events) 1

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Question 1 of 5

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)

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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

$1/2$

$(a) 1/4$

$(b) 1/2$

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)
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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

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)

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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 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

$3/8$

$(a) 1/8$

$(b) 3/8$

Question 4 of 5

Find the probability of tossing a normal coin thrice and getting:

$(a) 3$ of a kind

$(b)$ at least

$2$ Heads

Write fractions in the format “a/b”

$(a)$ (¼, 1/4, 2/8)

$(b)$ (½, 1/2, 4/8)

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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

$3$ 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

$1/2$

$(a) 1/4$

$(b) 1/2$

Question 5 of 5

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)

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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

$\mathsf{P(A\:and\:B)}=\mathsf{P(A)}\times\mathsf{P(B)}$

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

$1/4$

$1/4$