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)

$(ii)$ (¾, 3/4, 0.75)

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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:

$(a) 1$

$(b) 3$

$(c) 4$ or less

Write fractions in the format “a/b”

$(a)$ (⅓, 1/3, 2/6, 0.33)

$(b)$ (1/6, 0.17)

$(c)$ (5/6, 0.83)

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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.

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

$1$ .

favourable outcomes

$=$ $2$ (

$1,1$ )

total outcomes

$=$ $6$ (

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

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

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

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

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

$3$ .

favourable outcomes

$=$ $1$ (

$3$ )

total outcomes

$=$ $6$ (

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

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

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

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

$4$ or less.

favourable outcomes

$=$ $5$ (

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

total outcomes

$=$ $6$ (

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

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

	$=$	$\frac{\color{#e65021}{5}}{\color{#007DDC}{6}}$	Substitute values

$(a) 1/3$ or

$2/6$

$(b) 1/6$

$(c) 5/6$

Question 3 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)

$(e)$ (⅔, 2/3, 4/6, 0.67)

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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 4 of 7

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

$(a) 5$

$(b)$ an Odd number of dots

$(c)$ an Even number of dots

Write fractions in the format “a/b”

$(a)$ (1/6, 0.17)

$(b)$ (½, 1/2, 3/6, 0.5)

$(c)$ (½, 1/2, 3/6, 0.5)

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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.

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

$5$ .

favourable outcomes

$=$ $1$ (

$5$ )

total outcomes

$=$ $6$ (

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

$\mathsf{P(}5\mathsf{)}$	$=$	$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$	Probability Formula

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

$(b)$ Find the probability of the rolling the dice and getting an Odd number of dots.

favourable outcomes

$=$ $3$ (

$1,3,5$ )

total outcomes

$=$ $6$ (

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

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

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

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

$(c)$ Find the probability of the rolling the dice and getting an Even number of dots.

favourable outcomes

$=$ $3$ (

$2,4,6$ )

total outcomes

$=$ $6$ (

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

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

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

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

$(a) 1/6$

$(b) 1/2$

$(c) 1/2$

Question 5 of 7

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

$(d) 3$ or

$6$

$(e) gt 3$

$(f) lt 7$

Write fractions in the format “a/b”

$(d)$ (⅓, 1/3, 2/6, 0.17)

$(e)$ (½, 1/2, 3/6, 0.5)

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

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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 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 6 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:

$(a)$ Red

$(b)$ Yellow

Write fractions in the format “a/b”

$(a)$ (5/14, 0.36)

$(b)$ (3/7, 6/14, 0.43)

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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 drawing a marble at random and getting a Red marble.

favourable outcomes

$=$ $5$ (

$5$ Red)

total outcomes

$=$ $14$ (

$5$ Red,

$6$ Yellow,

$3$ Black)

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

	$=$	$\frac{\color{#e65021}{5}}{\color{#007DDC}{14}}$	Substitute values

$(b)$ Find the probability of the drawing a marble at random and getting a Yellow marble.

favourable outcomes

$=$ $6$ (

$6$ Red)

total outcomes

$=$ $14$ (

$5$ Red,

$6$ Yellow,

$3$ Black)

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

	$=$	$\frac{\color{#e65021}{6}}{\color{#007DDC}{14}}$	Substitute values

	$=$	$\frac{3}{7}$

$(a) 5/14$

$(b) 3/7$

Question 7 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)

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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.

$(c)$ 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

	$=$	$0$

$(c) 4/7$

$(d) 0$

Simple Probability 3

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