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

$4$

Write fractions in the format “a/b”

(1/36)

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Probability Formula

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

Product Rule

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

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

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 tossing a coin and getting Heads

favourable outcomes

$=$ $1$ (a coin has

$1$ Heads side)

total outcomes

$=$ $2$ (a coin has

$2$ 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

$4$

favourable outcomes

$=$ $1$ (a dice has

$1$ side with

$4$ )

total outcomes

$=$ $6$ (a dice has

$6$ 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

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

$5/108$

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Quiz summary

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1. Question

Hint

Great Work!

Probability Formula

Product Rule

2. Question

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Probability Formula

3. Question

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Fantastic!

Probability Formula

4. Question

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Correct!

Probability Formula

Addition Rule

Product Rule

5. Question

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Probability Formula

Product Rule

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