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Probability Tree (Independent Events)

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Probability Tree (Independent)

Probability Tree (Independent)

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Question 1 of 6

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

(a) 2

$(a) 2$ Tails

(b)

$(b)$ Heads and Tails (in any order)

Write fractions in the format “a/b”

$(a)$ $(a)$ (1/4)

$(b)$ $(b)$ (1/2, 2/4)

Incorrect

Help Video

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 6

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”

$(a)$ (1/8)

$(b)$ (3/8)

Incorrect

Help Video

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 3 of 6

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”

$(a)$ (1/4, 2/8)

$(b)$ (1/2, 4/8)

Incorrect

Help Video

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 4 of 6

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”

(1/4, 2/8)

Incorrect

Help Video

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

$1/4$

$1/4$

Question 5 of 6

A couple plan to have three kids. Find the probability of having:

$(a) 2$ Girls and

$1$ Boy (in any order)

$(b) 3$ of the same gender

Write fractions in the format “a/b”

$(a)$ (3/8)

$(b)$ (1/4, 2/8)

Incorrect

Help Video

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 having

$2$ Girls and

$1$ Boy.

First, set up a probability tree showing all possible genders of

$3$ kids

Find the probability of all outcomes

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

	$=$	$\frac{\color{#e65021}{1}}{\color{#007DDC}{2}}$	$1$ out of $2$ possible genders

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

	$=$	$\frac{\color{#e65021}{1}}{\color{#007DDC}{2}}$	$1$ out of $2$ possible genders

Mark each outcome with its corresponding probability

Pick the branches that lead to

$2$ Girls and

$1$ Boy

First branch (BGG):

$1/2$

$times$ $1/2$

$times$ $1/2$

$=$

$1/8$

Product Rule

Second branch (GBG):

$1/2$

$times$ $1/2$

$times$ $1/2$

$=$

$1/8$

Product Rule

Third branch (GGB):

$1/2$

$times$ $1/2$

$times$ $1/2$

$=$

$1/8$

Product Rule

Finally, add the solved probability for each branch

$1/8$

$+$ $1/8$

$+$ $1/8$

$=$

$3/8$

Addition Rule

Therefore, the probability of having

$2$ Girls and

$1$ Boy is

$3/8$

$(b)$ Find the probability of having

$3$ of the same sex.

Use the probability tree from part

$(a)$ and pick the branches that lead to

$3$ of the same sex then multiply the probabilities along it

First branch (BBB):

$1/2$

$times$ $1/2$

$times$ $1/2$

$=$

$1/8$

Product Rule

Second branch (GGG):

$1/2$

$times$ $1/2$

$times$ $1/2$

$=$

$1/8$

Product Rule

Finally, add the solved probability for each branch

$1/8$

$+$ $1/8$

$=$

$2/8$

$=$

$1/4$

Addition Rule

Therefore, the probability of having

$3$ of the same sex is

$1/4$

$(a) 3/8$

$(b) 1/4$

Question 6 of 6

A box contains

$5$ balls,

$3$ of which are Yellow and

$2$ are Blue. Find the probability of drawing

$3$ balls at random and getting:

$(a) 3$ Yellow

$(b)$ at least

$1$ Blue

Write fractions in the format “a/b”

$(a)$ (27/125)

$(b)$ (98/125)

Incorrect

Help Video

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

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

$(a)$ Find the probability of drawing

$3$ Yellow balls.

First, set up a probability tree showing all possible outcomes of drawing

$3$ balls

Find the probability of all outcomes

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

	$=$	$\frac{\color{#e65021}{3}}{\color{#007DDC}{5}}$	$3$ Yellow balls out of $5$ balls

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

	$=$	$\frac{\color{#e65021}{2}}{\color{#007DDC}{5}}$	$2$ Blue balls out of $5$ balls

Mark each outcome with its corresponding probability

Finally, pick the branch that lead to

$3$ Yellow and multiply the probabilities along it

$3/5$

$times$ $3/5$

$times$ $3/5$

$=$

$27/125$

Product Rule

Therefore, the probability of drawing at

$3$ Yellow balls is

$27/125$

$(b)$ Find the probability of drawing at least

$1$ Blue ball.

Notice that all outcomes have at least

$1$ Blue except YYY

Hence, we can simply get the complement of P(YYY)

$\mathsf{P(\dot{YYY})}$	$=$	$1-\mathsf{P(YYY)}$	Complementary Probability

	$=$	$1-27/125$	Substitute value

	$=$	$125/125-27/125$

	$=$	$98/125$

Therefore, the probability of drawing at least

$1$ Blue is

$98/125$

$(a) 27/125$

$(b) 98/125$