Independent Events 1

Question 1 of 5

A multi-stage event includes three stages: spinning a spinner, tossing a coin, and rolling a dice. Find the probability of getting Yellow, Heads and the side

4

$4$

Write fractions in the format “a/b”

(1/36)

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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}

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

Product Rule

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

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

Find the probability of the arrow landing on Yellow when spinning the spinner

favourable outcomes

=

$=$ $1$ $1$ (

1

$1$ Yellow section)

total outcomes

=

$=$ $3$ $3$ (

3

$3$ total sections)

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

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

Find the probability of tossing a coin and getting Heads

favourable outcomes

=

$=$ $1$ $1$ (a coin has

1

$1$ Heads side)

total outcomes

=

$=$ $2$ $2$ (a coin has

2

$2$ sides)

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

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

Find the probability of rolling a dice and getting

4

$4$

favourable outcomes

=

$=$ $1$ $1$ (a dice has

1

$1$ side with

4

$4$ )

total outcomes

=

$=$ $6$ $6$ (a dice has

6

$6$ sides)

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

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

Now, substitute the probabilities into the product rule

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

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

$\mathsf{P(4)}$	$=$ $=$	$\frac{1}{6}$

$\mathsf{P(Yellow\:and\:Head\:and\:4)}$	$=$ $=$	$\mathsf{P(Yellow)}\times\mathsf{P(Heads)}\times\mathsf{P(4)}$	Product Rule

	$=$	$\frac{1}{3}\times\frac{1}{2}\times\frac{1}{6}$	Substitute values

	$=$	$\frac{1}{36}$

$1/36$

Question 2 of 5

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:

$(a)$ Yellow and Green

$(b)$ Blue and Orange

Write fractions in the format “a/b”

$(a)$ (1/18)

$(b)$ (1/9, 2/18)

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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 Yellow and Green.

Find the probability of the arrow landing on Yellow

favourable outcomes

$=$ $1$ (

$1$ Yellow section)

total outcomes

$=$ $3$ (

$3$ total sections)

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

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

Find the probability of the arrow landing on Green

favourable outcomes

$=$ $1$ (

$1$ Green section)

total outcomes

$=$ $6$ (

$6$ total sections)

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

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

Now, substitute the probabilities into the product rule

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

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

$\mathsf{P(Yellow\:and\:Green)}$	$=$	$\mathsf{P(Yellow)}\times\mathsf{P(Green)}$	Product Rule

	$=$	$\frac{1}{3}\times\frac{1}{6}$	Substitute values

	$=$	$\frac{1}{18}$

Therefore, the probability of the arrow landing on Yellow and Green is

$1/18$ .

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

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 Orange

favourable outcomes

$=$ $2$ (

$2$ Orange sections)

total outcomes

$=$ $6$ (

$6$ total sections)

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

	$=$	$\frac{\color{#e65021}{2}}{\color{#007DDC}{6}}$	Substitute values

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

Now, substitute the probabilities into the product rule

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

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

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

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

	$=$	$\frac{1}{9}$

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

$1/9$ .

$(a) 1/18$

$(b) 1/9$

Question 3 of 5

A coin is weighted in such a way that Tails would show up twice the chance of Heads. Find the probability of tossing this coin thrice and getting:

$(a) 3$ Tails

$(b)$ Tails, Heads, Tails

Write fractions in the format “a/b”

$(a)$ (8/27)

$(b)$ (4/27)

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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 getting

$3$ Tails.

Find the probability of getting Tails. Remember that Tails has twice the chance as Heads

favourable outcomes

$=$ $2$ (T,T)

total outcomes

$=$ $3$ (H,T,T)

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

	$=$	$\frac{\color{#e65021}{2}}{\color{#007DDC}{3}}$	Substitute values

Now, substitute this probability into the product rule

$\mathsf{P(3\:Tails)}$	$=$	$\mathsf{P(Tails)}\times\mathsf{P(Tails)}\times\mathsf{P(Tails)}$	Product Rule

	$=$	$\frac{2}{3}\times\frac{2}{3}\times\frac{2}{3}$	Substitute values

	$=$	$\frac{8}{27}$

Therefore, the probability of getting

$3$ Tails is

$8/27$ .

$(b)$ Find the probability of getting Tails, Heads, Tails.

From part

$(a)$ , we have solved for the probability of getting Tails

$\mathsf{P(Tails)}$

$=$

$\frac{2}{3}$

Find the probability of getting Heads. Remember that Tails has twice the chance as Heads

favourable outcomes

$=$ $1$ (H)

total outcomes

$=$ $3$ (H,T,T)

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

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

Now, substitute the solved probabilities into the product rule

$\mathsf{P(Tails)}$	$=$	$\frac{2}{3}$

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

$\mathsf{P(Tails,\:Heads,\:Tails)}$	$=$	$\mathsf{P(Tails)}\times\mathsf{P(Heads)}\times\mathsf{P(Tails)}$	Product Rule

	$=$	$\frac{2}{3}\times\frac{1}{3}\times\frac{2}{3}$	Substitute values

	$=$	$\frac{4}{27}$

Therefore, the probability of getting Tails, Heads, Tails is

$4/27$ .

$(a) 8/27$

$(b) 4/27$

Question 4 of 5

One box contains cards labelled

$1$ to

$5$ and another box contains cards labelled

$A$ to

$E$ . If you draw a card from each box, find the probability of drawing:

$(a)$ A

$2$ or

$4$ , then an

$A$

$(b)$ An Even Number and a Vowel

Write fractions in the format “a/b”

$(a)$ (2/25)

$(b)$ (4/25)

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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 a

$2$ or

$4$ then an

$A$ .

Find the probability of getting

$2$ or

$4$ .

favourable outcomes

$=$ $2$ (2,4)

total outcomes

$=$ $5$ (5 number cards)

$\mathsf{P(2\:or\:4)}$	$=$	$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$	Probability Formula

	$=$	$\frac{\color{#e65021}{2}}{\color{#007DDC}{5}}$	Substitute values

Find the probability of getting

$A$ .

favourable outcomes

$=$ $1$ (A)

total outcomes

$=$ $5$ (5 letter cards)

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

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

Now, substitute this probability into the product rule

$\mathsf{P(3\:Tails)}$	$=$	$\mathsf{P(2\:or\:4)}\times\mathsf{P(A)}$	Product Rule

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

	$=$	$\frac{2}{25}$

Therefore, the probability of getting a

$2$ or a

$4$ then an

$A$ is

$2/25$ .

$(b)$ Find the probability of getting an Even Number and a Vowel.

Find the probability of getting an Even Number,

$2$ or

$4$ .

favourable outcomes

$=$ $2$ (2,4)

total outcomes

$=$ $5$ (5 number cards)

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

	$=$	$\frac{\color{#e65021}{2}}{\color{#007DDC}{5}}$	Substitute values

Find the probability of getting a Vowel,

$A$ or

$E$ .

favourable outcomes

$=$ $2$ (A,E)

total outcomes

$=$ $5$ (5 letter cards)

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

	$=$	$\frac{\color{#e65021}{2}}{\color{#007DDC}{5}}$	Substitute values

Now, substitute this probability into the product rule

$\mathsf{P(3\:Tails)}$	$=$	$\mathsf{P(Even\:Number)}\times\mathsf{P(Vowel)}$	Product Rule

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

	$=$	$\frac{4}{25}$

Therefore, the probability of getting an Even Number and a Vowel is

$4/25$ .

$(a) 2/25$

$(b) 4/25$

Question 5 of 5

Find the probability of rolling two dice and getting double numbers and a sum of at least

$9$

Write fractions in the format “a/b”

(5/108, 10/216)

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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 double number

favourable outcomes

$=$ $6$ (a dice has

$6$ numbers hence

$6$ possible double numbers)

total outcomes

$=$ $36$

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

	$=$	$\frac{\color{#e65021}{6}}{\color{#007DDC}{36}}$	Substitute values

	$=$	$\frac{1}{6}$

Find the probability of getting a sum of at least

$9$

favourable outcomes

$=$ $10$ (

$10$ dots on lattice above)

total outcomes

$=$ $36$

$\mathsf{P(sum}≤9)$	$=$	$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$	Probability Formula

	$=$	$\frac{\color{#e65021}{10}}{\color{#007DDC}{36}}$	Substitute values

	$=$	$\frac{5}{18}$

Finally, multiply the two probabilities

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

$\mathsf{P(sum}≤9)$	$=$	$\frac{5}{8}$

$\mathsf{P(double\:and\:sum}≤9)$	$=$	$\mathsf{P(double)}\times\mathsf{P(sum}≤9)$	Product Rule

	$=$	$\frac{1}{6}\times\frac{5}{18}$	Substitute values

	$=$	$\frac{5}{108}$

$5/108$

Independent Events 1

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Quiz summary

Information

1. Question

Hint

Great Work!

Probability Formula

Product Rule

2. Question

Hint

Nice Job!

Probability Formula

3. Question

Hint

Fantastic!

Probability Formula

4. Question

Hint

Correct!

Probability Formula

Addition Rule

Product Rule

5. Question

Hint

Great Work!

Probability Formula

Product Rule

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