Independent Events 2

Question 1 of 4

Two spinners are spun one after the other. Find the probability of getting the following outcomes from the

1

$1$ st and

2

$2$ nd spins respectively:

(a)

$(a)$ Orange and Red

(b)

$(b)$ Blue and any color

Write fractions in the format “a/b”

$(a)$ $(a)$ (1/12)

$(b)$ $(b)$ (⅓, 1/3, 4/12)

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Probability Formula

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

$(a)$ Find the probability of the arrow landing on Orange and Red.

Find the probability of the arrow landing on Orange

<

favourable outcomes

$=$ $1$ (

$1$ Orange section)

total outcomes

$=$ $3$ (

$3$ total sections)

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

	$=$	$\frac{\color{#e65021}{1}}{\color{#007DDC}{3}}$	Substitute values

Find the probability of the arrow landing on Red

favourable outcomes

$=$ $1$ (

$1$ Red section)

total outcomes

$=$ $4$ (

$4$ total sections)

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

	$=$	$\frac{\color{#e65021}{1}}{\color{#007DDC}{4}}$	Substitute values

Now, substitute the probabilities into the product rule

$\mathsf{P(Orange)}$	$=$	$\frac{1}{3}$

$\mathsf{P(Red)}$	$=$	$\frac{1}{4}$

$\mathsf{P(Orange\:and\:Red)}$	$=$	$\mathsf{P(Orange)}\times\mathsf{P(Red)}$	Product Rule

	$=$	$\frac{1}{3}\times\frac{1}{4}$	Substitute values

	$=$	$\frac{1}{12}$

Therefore, the probability of the arrow landing on Orange and Red is

$1/12$ .

$(b)$ Find the probability of the arrow landing on Blue and any color.

Find the probability of the arrow landing on Blue

favourable outcomes

$=$ $1$ (

$1$ Blue section)

total outcomes

$=$ $3$ (

$3$ total sections)

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

	$=$	$\frac{\color{#e65021}{1}}{\color{#007DDC}{3}}$	Substitute values

Find the probability of the arrow landing on any color

favourable outcomes

$=$ $4$ (any color or section)

total outcomes

$=$ $4$ (

$4$ total sections)

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

	$=$	$\frac{\color{#e65021}{4}}{\color{#007DDC}{4}}$	Substitute values

	$=$	$1$

Now, substitute the probabilities into the product rule

$\mathsf{P(Blue)}$	$=$	$\frac{1}{3}$

$\mathsf{P(any\:color)}$	$=$	$1$

$\mathsf{P(Blue\:and\:any\:color)}$	$=$	$\mathsf{P(Blue)}\times\mathsf{P(any\:color)}$	Product Rule

	$=$	$\frac{1}{3}\times1$	Substitute values

	$=$	$\frac{1}{3}$

Therefore, the probability of the arrow landing on Blue and any color is

$1/3$ .

$(a) 1/12$

$(b) 1/3$

Question 2 of 4

Two spinners are spun one after the other. Find the probability of getting the following outcomes from the

$1$ st and

$2$ nd spins respectively:

$(a)$ Orange and Black

$(b)$ Pink and red

Write fractions in the format “a/b”

$(a)$ (1/30)

$(b)$ (1/10, 3/30)

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Probability Formula

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

$(a)$ Find the probability of the arrow landing on Orange and Black.

Find the probability of the arrow landing on Orange

favourable outcomes

$=$ $1$ (

$1$ Orange section)

total outcomes

$=$ $5$ (

$5$ total sections)

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

	$=$	$\frac{\color{#e65021}{1}}{\color{#007DDC}{5}}$	Substitute values

Find the probability of the arrow landing on Black

favourable outcomes

$=$ $1$ (

$3$ Black section)

total outcomes

$=$ $6$ (

$6$ total sections)

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

	$=$	$\frac{\color{#e65021}{1}}{\color{#007DDC}{6}}$	Substitute values

Now, substitute the probabilities into the product rule

$\mathsf{P(Orange)}$	$=$	$\frac{1}{5}$

$\mathsf{P(Black)}$	$=$	$\frac{1}{6}$

$\mathsf{P(Orange\:and\:Black)}$	$=$	$\mathsf{P(Orange)}\times\mathsf{P(Black)}$	Product Rule

	$=$	$\frac{1}{5}\times\frac{1}{6}$	Substitute values

	$=$	$\frac{1}{30}$

Therefore, the probability of the arrow landing on Orange and Red is

$1/30$ .

$(b)$ Find the probability of the arrow landing on Pink and Red.

Find the probability of the arrow landing on Pink

favourable outcomes

$=$ $1$ (

$1$ Pink section)

total outcomes

$=$ $5$ (

$5$ total sections)

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

	$=$	$\frac{\color{#e65021}{1}}{\color{#007DDC}{5}}$	Substitute values

Find the probability of the arrow landing on Red

favourable outcomes

$=$ $3$ (

$3$ Red sections)

total outcomes

$=$ $6$ (

$6$ total sections)

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

	$=$	$\frac{\color{#e65021}{3}}{\color{#007DDC}{6}}$	Substitute values

	$=$	$\frac{1}{2}$

Now, substitute the probabilities into the product rule

$\mathsf{P(Pink)}$	$=$	$\frac{1}{5}$

$\mathsf{P(Red)}$	$=$	$\frac{1}{2}$

$\mathsf{P(Pink\:and\:Red)}$	$=$	$\mathsf{P(Pink)}\times\mathsf{P(Red)}$	Product Rule

	$=$	$\frac{1}{5}\times\frac{1}{2}$	Substitute values

	$=$	$\frac{1}{10}$

Therefore, the probability of the arrow landing on Pink and Red is

$1/10$ .

$(a) 1/30$

$(b) 1/10$

Question 3 of 4

$4$ coins were tossed at the same time. Find the probability that the

$2nd$ and

$3rd$ coins will be tails.

Write fractions in the format “a/b”

(¼, 1/4, 4/16)

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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, find the probability of getting Tails and getting Heads.

favourable outcomes

$=$ $1$ (T)

total outcomes

$=$ $2$ (H,T)

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

	$=$	$\frac{\color{#e65021}{1}}{\color{#007DDC}{2}}$	Substitute values

favourable outcomes

$=$ $1$ (H)

total outcomes

$=$ $2$ (H,T)

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

	$=$	$\frac{\color{#e65021}{1}}{\color{#007DDC}{2}}$	Substitute values

There are four probabilities where the second and third coin would come up as Tails.

$\text(H)$ $\text(TT)$ $\text(H)$		$\text(H)$ $\text(TT)$ $\text(T)$
$\text(T)$ $\text(TT)$ $\text(H)$		$\text(T)$ $\text(TT)$ $\text(T)$

Substitute the probability of getting Heads or Tails to each outcome.

$\mathsf{P(HTTH)}$	$=$	$\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}$

	$=$	$\frac{1}{16}$

$\mathsf{P(HTTT)}$	$=$	$\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}$

	$=$	$\frac{1}{16}$

$\mathsf{P(TTTH)}$	$=$	$\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}$

	$=$	$\frac{1}{16}$

$\mathsf{P(TTTT)}$	$=$	$\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}$

	$=$	$\frac{1}{16}$

Finally, use the addition rule to get the total probability.

$\mathsf{P(A\:or\:B)}$	$=$	$\mathsf{P(A)}+\mathsf{P(B)}$	Addition Rule
	$=$	$\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}$	Substitute values
	$=$	$\frac{4}{16}$
	$=$	$\frac{1}{4}$	Simplify

$1/4$

Question 4 of 4

Find the probability of rolling two dice and getting a sum of

$4$ and a sum greater than

$10$

Write fractions in the format “a/b”

(1/144, 3/432)

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Probability Formula

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

Product Rule

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

First, set up a lattice showing all possible sums for the two dice

This means there are $36$ possible outcomes

Next, find the probability of getting a sum of

$4$

favourable outcomes

$=$ $3$ (

$3$ dots on lattice above)

total outcomes

$=$ $36$

$\mathsf{P(4)}$	$=$	$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$	Probability Formula

	$=$	$\frac{\color{#e65021}{3}}{\color{#007DDC}{36}}$	Substitute values

	$=$	$\frac{1}{12}$

Find the probability of getting a sum greater than

$10$

favourable outcomes

$=$ $3$ (

$3$ dots on lattice above)

total outcomes

$=$ $36$

$\mathsf{P(sum}>10)$	$=$	$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$	Probability Formula

	$=$	$\frac{\color{#e65021}{3}}{\color{#007DDC}{36}}$	Substitute values

	$=$	$\frac{1}{12}$

Finally, multiply the two probabilities

$\mathsf{P(4)}$	$=$	$\frac{1}{12}$

$\mathsf{P(sum}>10)$	$=$	$\frac{1}{12}$

$\mathsf{P(4\:and\:sum}>10)$	$=$	$\mathsf{P(4)}\times\mathsf{P(sum}>10)$	Product Rule

	$=$	$\frac{1}{12}\times\frac{1}{12}$	Substitute values

	$=$	$\frac{1}{144}$

$1/144$

Independent Events 2

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Quiz summary

Information

1. Question

Hint

Well Done!

Probability Formula

2. Question

Hint

Excellent!

Probability Formula

3. Question

Hint

Keep Going!

Probability Formula

Addition Rule

Product Rule

4. Question

Hint

Keep Going!

Probability Formula

Product Rule

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