Probability Tree (Independent Events) 2

Question 1 of 5

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

(a) 2

$(a) 2$ Girls and

1

$1$ Boy (in any order)

(b) 3

$(b) 3$ of the same gender

Write fractions in the format “a/b”

$(a)$ $(a)$ (⅜, 3/8)

$(b)$ $(b)$ (¼, 1/4, 2/8)

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Probability Formula

P (E) = \frac{f a v o u r a b l e o u t c o m e s}{t o t a l o u t c o m e s}

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

Addition Rule

P (A o r B) = P (A) + P (B)

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

Product Rule

P (A a n d B) = P (A) \times P (B)

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

(a)

$(a)$ Find the probability of having

2

$2$ Girls and

1

$1$ Boy.

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

3

$3$ kids

Find the probability of all outcomes

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

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

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

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

Mark each outcome with its corresponding probability

Pick the branches that lead to

2

$2$ Girls and

1

$1$ Boy

First branch (BGG):

$\frac{1}{2}$ $1/2$

\times

$times$ $\frac{1}{2}$ $1/2$

\times

$times$ $\frac{1}{2}$ $1/2$

=

$=$

\frac{1}{8}

$1/8$

Product Rule

Second branch (GBG):

$\frac{1}{2}$ $1/2$

\times

$times$ $\frac{1}{2}$ $1/2$

\times

$times$ $\frac{1}{2}$ $1/2$

=

$=$

\frac{1}{8}

$1/8$

Product Rule

Third branch (GGB):

$\frac{1}{2}$ $1/2$

\times

$times$ $\frac{1}{2}$ $1/2$

\times

$times$ $\frac{1}{2}$ $1/2$

=

$=$

\frac{1}{8}

$1/8$

Product Rule

Finally, add the solved probability for each branch

$\frac{1}{8}$ $1/8$

+

$+$ $\frac{1}{8}$ $1/8$

+

$+$ $\frac{1}{8}$ $1/8$

=

$=$

\frac{3}{8}

$3/8$

Addition Rule

Therefore, the probability of having

2

$2$ Girls and

1

$1$ Boy is

\frac{3}{8}

$3/8$

(b)

$(b)$ Find the probability of having

3

$3$ of the same gender.

Use the probability tree from part

(a)

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

3

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

First branch (BBB):

$\frac{1}{2}$ $1/2$

\times

$times$ $\frac{1}{2}$ $1/2$

\times

$times$ $\frac{1}{2}$ $1/2$

=

$=$

\frac{1}{8}

$1/8$

Product Rule

Second branch (GGG):

$\frac{1}{2}$ $1/2$

\times

$times$ $\frac{1}{2}$ $1/2$

\times

$times$ $\frac{1}{2}$ $1/2$

=

$=$

\frac{1}{8}

$1/8$

Product Rule

Finally, add the solved probability for each branch

$\frac{1}{8}$ $1/8$

+

$+$ $\frac{1}{8}$ $1/8$

=

$=$

\frac{2}{8}

$2/8$

=

$=$

\frac{1}{4}

$1/4$

Addition Rule

Therefore, the probability of having

3

$3$ of the same gender is

\frac{1}{4}

$1/4$

(a) \frac{3}{8}

$(a) 3/8$

(b) \frac{1}{4}

$(b) 1/4$

Question 2 of 5

A box contains

5

$5$ balls,

3

$3$ of which are Yellow and

2

$2$ are Blue. Find the probability of drawing

3

$3$ balls at random and getting:

(a) 3

$(a) 3$ Yellow

$(b)$ at least

$1$ Blue

Write fractions in the format “a/b”

$(a)$ (27/125)

$(b)$ (98/125)

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