Simple Probability 3

Question 1 of 7

Find the probability of spinning the following spinners and getting:

$(i)$ Pink or Green

$(ii)$ Yellow or Blue

Write fractions in the format “a/b”

$(i)$ (2/5, 0.4)

$(ii)$ (3/4, 0.75)

Incorrect

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Probability Formula

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

Addition Principle

If two or more events are joined by OR, their probabilities should be added.

$(i)$ Find the probability of the arrow landing on Pink or Green.

favourable outcomes

$=$ $2$ (

$1$ Pink

$+1$ Green)

total outcomes

$=$ $5$ (Pink, Green, Orange, Blue, Yellow)

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

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

$(ii)$ Find the probability of the arrow landing on Yellow or Blue.

favourable outcomes

$=$ $3$ (

$2$ Yellow

$+1$ Blue)

total outcomes

$=$ $4$ (Orange, Blue, and

$2$ Yellow)

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

	$=$	$\frac{\color{#e65021}{3}}{\color{#007DDC}{4}}$	Substitute values

$(i) 2/5$

$(ii) 3/4$

Question 2 of 7

A six-sided dice does not have a side with

$6$ dots but instead has two sides with

$1$ dot. Find the probability of the rolling the dice and getting:

$(d) 6$

$(e)$ greater than

$1$

Write fractions in the format “a/b”

$(d)$ (0, 0/6)

$(e)$ (2/3, 4/6, 0.67)

Incorrect

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Probability Formula

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

Addition Principle

If two or more events are joined by OR, their probabilities should be added.

$(d)$ Find the probability of rolling the dice and getting

$6$ .

favourable outcomes

$=$ $0$ (no sides have

$6$ dots)

total outcomes

$=$ $6$ (

$1,1,2,3,4,5$ )

$\mathsf{P(}6\mathsf{)}$	$=$	$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$	Probability Formula

	$=$	$\frac{\color{#e65021}{0}}{\color{#007DDC}{6}}$	Substitute values

	$=$	$0$

$(e)$ Find the probability of rolling the dice and getting greater than

$1$ .

favourable outcomes

$=$ $4$ (

$2,3,4,5$ )

total outcomes

$=$ $6$ (

$1,1,2,3,4,5$ )

$\mathsf{P(greater\:than\:}1\mathsf{)}$	$=$	$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$	Probability Formula

	$=$	$\frac{\color{#e65021}{4}}{\color{#007DDC}{6}}$	Substitute values

	$=$	$\frac{2}{3}$

$(d) 0$

$(e) 4/6$ or

$2/3$

Question 3 of 7

A normal six-sided dice is rolled. Find the probability of getting:

$(d) 3$ or

$6$

$(e) “>“3$

$(f) “<“7$

Write fractions in the format “a/b”

$(d)$ (1/3, 2/6, 0.17)

$(e)$ (1/2, 3/6, 0.5)

$(f)$ (1, 6/6)

Incorrect

Help Video

Probability Formula

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

Addition Principle

If two or more events are joined by OR, their probabilities should be added.

$(d)$ Find the probability of the rolling the dice and getting

$3$ or

$6$ .

favourable outcomes

$=$ $2$ (

$3,6$ )

total outcomes

$=$ $6$ (

$1,2,3,4,5,6$ )

$\mathsf{P(}3\:\mathsf{or}\:6\mathsf{)}$	$=$	$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$	Probability Formula

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

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

$(e)$ Find the probability of the rolling the dice and getting

$>“3$ .

favourable outcomes

$=$ $3$ (

$4,5,6$ )

total outcomes

$=$ $6$ (

$1,2,3,4,5,6$ )

$\mathsf{P(}>3\mathsf{)}$	$=$	$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$	Probability Formula

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

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

$(f)$ Find the probability of the rolling the dice and getting

$<“7$ .

favourable outcomes

$=$ $6$ (

$1,2,3,4,5,6$ )

total outcomes

$=$ $6$ (

$1,2,3,4,5,6$ )

$\mathsf{P(}<7\mathsf{)}$	$=$	$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$	Probability Formula

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

	$=$	$1$

$(d) 1/3$

$(e) 1/2$

$(f) 1$

Question 4 of 7

A jar contains

$5$ red marbles,

$6$ yellow marbles and

$3$ black marbles. Find the probability of drawing a marble at random and getting:

$(c)$ Black or Red

$(d)$ Blue

Write fractions in the format “a/b”

$(c)$ (4/7, 8/14, 0.57)

$(d)$ (0, 0/14)

Incorrect

Help Video

Probability Formula

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

Addition Principle

If two or more events are joined by OR, their probabilities should be added.

$(c)$ Find the probability of the drawing a marble at random and getting a Black or Red marble.

favourable outcomes

$=$ $8$ (

$3$ Black,

$5$ Red)

total outcomes

$=$ $14$ (

$5$ Red,

$6$ Yellow,

$3$ Black)

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

	$=$	$\frac{\color{#e65021}{8}}{\color{#007DDC}{14}}$	Substitute values

	$=$	$\frac{4}{7}$

$(d)$ Find the probability of the drawing a marble at random and getting a Blue marble.

favourable outcomes

$=$ $0$ (there are no Blue marbles)

total outcomes

$=$ $14$ (

$5$ Red,

$6$ Yellow,

$3$ Black)

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

	$=$	$\frac{\color{#e65021}{0}}{\color{#007DDC}{14}}$	Substitute values

	$=$	$0$

$(c) 4/7$

$(d) 0$

Question 5 of 7

Find the probability of drawing from a standard deck of cards and getting:

$(c)$ Picture

$(d)$ Red or Black

Write fractions in the format “a/b”

$(c)$ (3/13, 12/52, 0.23)

$(d)$ (1, 52/52)

Incorrect

Help Video

Probability Formula

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

Addition Principle

If two or more events are joined by OR, their probabilities should be added.

$(c)$ Find the probability of drawing from a standard deck of cards and getting a Picture card.

favourable outcomes

$=3*4=$ $12$ (a standard deck has

$3$ Picture cards for each of the

$4$ suits)

total outcomes

$=$ $52$ (a standard deck has

$52$ cards)

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

	$=$	$\frac{\color{#e65021}{12}}{\color{#007DDC}{52}}$	Substitute values

	$=$	$\frac{3}{13}$

$(d)$ Find the probability of drawing from a standard deck of cards and getting a Red or Black card.

favourable outcomes

$=$ $52$ (all cards from a standard deck are either Red or Black)

total outcomes

$=$ $52$ (a standard deck has

$52$ cards)

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

	$=$	$\frac{\color{#e65021}{52}}{\color{#007DDC}{52}}$	Substitute values

	$=$	$1$

$(c) 3/13$

$(d) 1$

Question 6 of 7

Find the probability of drawing from a standard deck of cards and getting:

$(c)$ Red or a Black

$9$

$(d)$ Red

$4$

Write fractions in the format “a/b”

$(c)$ (7/13, 28/52, 0.52)

$(d)$ (2/52, 1/26, 0.04)

Incorrect

Help Video

Probability Formula

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

Addition Principle

If two or more events are joined by OR, their probabilities should be added.

$(c)$ Find the probability of drawing from a standard deck of cards and getting a Red or a Black

$9$ card.

favourable outcomes

$=26+1+1=$ $28$ (

$26$ Red, one

$9$ of Clubs, one

$9$ of Spades)

total outcomes

$=$ $52$ (a standard deck has

$52$ cards)

$\mathsf{P(Red\:or\:a\:Black\:}9\mathsf{)}$	$=$	$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$	Probability Formula

	$=$	$\frac{\color{#e65021}{28}}{\color{#007DDC}{52}}$	Substitute values

	$=$	$\frac{7}{13}$

$(d)$ Find the probability of drawing from a standard deck of cards and getting a Red

$4$ card.

favourable outcomes

$=$ $2$ (one

$4$ of Hearts, one

$4$ of Diamonds)

total outcomes

$=$ $52$ (a standard deck has

$52$ cards)

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

	$=$	$\frac{\color{#e65021}{2}}{\color{#007DDC}{52}}$	Substitute values

	$=$	$\frac{1}{26}$

$(c) 28/52$ or

$7/13$

$(d) 2/52$ or

$1/26$

Question 7 of 7

Find the probability of drawing from a standard deck of cards and getting:

$(c)$ Black Jack

$(d)$ Queen or

$3$ of Diamonds

Write fractions in the format “a/b”

$(c)$ (2/52, 1/26, 0.04)

$(d)$ (5/52, [0.10)

Incorrect

Help Video

Probability Formula

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

Addition Principle

If two or more events are joined by OR, their probabilities should be added.

$(c)$ Find the probability of drawing from a standard deck of cards and getting a Black Jack.

favourable outcomes

$=$ $2$ (

$1$ Jack of Spades,

$1$ Jack of Clubs)

total outcomes

$=$ $52$ (a standard deck has

$52$ cards)

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

	$=$	$\frac{\color{#e65021}{2}}{\color{#007DDC}{52}}$	Substitute values

	$=$	$\frac{1}{26}$

$(d)$ Find the probability of drawing from a standard deck of cards and getting a Queen or a

$3$ of Diamonds card.

favourable outcomes

$=(4*1)+1=$ $5$ (each of the

$4$ suits have

$1$ Queen card, one

$3$ of Diamonds)

total outcomes

$=$ $52$ (a standard deck has

$52$ cards)

$\mathsf{P(Queen\:or\:}3\:\mathsf{of\:Diamonds)}$	$=$	$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$	Probability Formula

	$=$	$\frac{\color{#e65021}{5}}{\color{#007DDC}{52}}$	Substitute values

$(c) 2/52$ or

$1/26$

$(d) 5/52$

Simple Probability 3

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Quiz summary

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1. Question

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Nice Job!

Probability Formula

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2. Question

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Fantastic!

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3. Question

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Correct!

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4. Question

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Exceptional!

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