Simple Probability 1

Question 1 of 7

Given that the range of probability of an event happening is

0 \leq P (E) \leq 1

$0\leq\mathsf{P(E)}\leq1$ , enter the equivalent probability for each keyword below:

(i)

$(i)$ Certain

(i i)

$(ii)$ Even Chance

(i i i)

$(iii)$ Impossible

Write fractions in the format “a/b”

$(i)$ $(i)$ Certain $=$ $=$ (1, 1/1, 100/100)

$(i i)$ $(ii)$ Even Chance $=$ $=$ (½, 1/2, 0.5, 50/100)

$(i i i)$ $(iii)$ Impossible $=$ $=$ (0, 0/1, 0/100)

Incorrect

Range of Probability

0 \leq P (E) \leq 1

$0\leq\mathsf{P(E)}\leq1$

(i)

$(i)$ Find the equivalent probability of being certain that an event will happen.

This also means a

100 %

$100%$ chance for an event to happen, which is equivalent to the highest possible value for a probability

certain

= 1

$=1$

(i i)

$(ii)$ Find the equivalent probability of having an even chance that an event will happen.

This also means

50 %

$50%$ chance for an even to happen. An example is tossing a coin since there is an even chance for Heads or Tails to show up.

even chance

= \frac{1}{2}

$=1/2$ or

0.5

$0.5$

(i i i)

$(iii)$ Find the equivalent probability for an event to be impossible.

This also means a

0 %

$0%$ chance for an event to happen, which is equivalent to the lowest possible value for a probability

impossible

= 0

$=0$

(i) 1

$(i) 1$

(i i) \frac{1}{2}

$(ii) 1/2$ or

0.5

$0.5$

(i i i) 0

$(iii) 0$

Question 2 of 7

Find the probability of tossing a normal coin and getting:

(i)

$(i)$ Heads

(i i)

$(ii)$ Tails

Write fractions in the format “a/b”

$(i)$ $(i)$ (½, 1/2, 50/100, 0.5, 5/10)

$(i i)$ $(ii)$ (½, 1/2, 50/100, 0.5, 5/10)

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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}

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

(i)

$(i)$ Find the probability of getting Heads from the coin toss.

favourable outcomes

=

$=$ $1$ $1$ (Heads)

total outcomes

=

$=$ $2$ $2$ (Heads, Tails)

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

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

(i i)

$(ii)$ Find the probability of getting Tails from the coin toss.

favourable outcomes

=

$=$ $1$ $1$ (Tails)

total outcomes

=

$=$ $2$ $2$ (Heads, Tails)

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

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

(i) \frac{1}{2}

$(i) 1/2$

(i i) \frac{1}{2}

$(ii) 1/2$

Question 3 of 7

Find the probability of spinning the following spinners and getting:

(i)

$(i)$ Orange

(i i)

$(ii)$ Pink

Write fractions in the format “a/b”

$(i)$ $(i)$ (½, 1/2, 2/4, 0.5)

$(i i)$ $(ii)$ (1/5, 0.2)

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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}

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

(i)

$(i)$ Find the probability of getting Orange from the spinner.

favourable outcomes

$=$ $2$ (Orange)

total outcomes

$=$ $4$ (Orange, Blue, Orange, Yellow)

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

	$=$	$\frac{\color{#e65021}{2}}{\color{#007DDC}{4}}$	Substitute values

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

$(ii)$ Find the probability of getting Pink from the spinner.

favourable outcomes

$=$ $1$ (Pink)

total outcomes

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

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

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

$(i) 1/2$

$(ii) 1/5$

Question 4 of 7

A spinning barrel holds

$100$ tickets for a raffle. Jake bought

$10$ of those tickets, Emma bought

$5$ , and Dylan bought

$1$ . Find the probability that:

$(a)$ Jake will win

$(b)$ Emma will win

Write fractions in the format “a/b”

$(a)$ (1/10, 10/100, 0.1)

$(b)$ (1/20, 5/100, 10/200, 0.05)

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Probability Formula

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

$(a)$ Find the probability that Jake will win.

favourable outcomes

$=$ $10$ (Jake bought

$10$ tickets)

total outcomes

$=$ $100$ (the barrel holds

$100$ tickets)

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

	$=$	$\frac{\color{#e65021}{10}}{\color{#007DDC}{100}}$	Substitute values

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

$(b)$ Find the probability that Emma will win.

favourable outcomes

$=$ $5$ (Emma bought

$5$ tickets)

total outcomes

$=$ $100$ (the barrel holds

$100$ tickets)

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

	$=$	$\frac{\color{#e65021}{5}}{\color{#007DDC}{100}}$	Substitute values

	$=$	$\frac{1}{20}$

$(a) 1/10$

$(b) 1/20$

Question 5 of 7

A spinning barrel holds

$100$ tickets for a raffle. Jake bought

$10$ of those tickets, Emma bought

$5$ , and Dylan bought

$1$ . Find the probability that:

$(c)$ Dylan will win

$(d)$ None of them will win

Write fractions in the format “a/b”

$(c)$ (1/100, 0.01)

$(d)$ (84/100, 21/25, 0.84)

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Probability Formula

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

$(c)$ Find the probability that Dylan will win.

favourable outcomes

$=$ $1$ (Dylan bought

$1$ ticket)

total outcomes

$=$ $100$ (the barrel holds

$100$ tickets)

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

	$=$	$\frac{\color{#e65021}{1}}{\color{#007DDC}{100}}$	Substitute values

$(d)$ Find the probability that none of them will win.

favourable outcomes

$=100-10-5-1=$ $84$ (tickets that were not bought by them)

total outcomes

$=$ $100$ (the barrel holds

$100$ tickets)

$\mathsf{P(none)}$	$=$	$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$	Probability Formula

	$=$	$\frac{\color{#e65021}{84}}{\color{#007DDC}{100}}$	Substitute values

	$=$	$\frac{21}{25}$

$(c) 1/100$

$(d) 84/100$ or

$21/25$

Question 6 of 7

A spinning wheel has

$16$ sectors marked

$1$ -

$16$ . Each of these sectors corresponds to a ticket that can be bought and a lady named Jackie wants to buy some. Answer the following:

$(a)$ How many tickets should she buy to have a certain chance at winning?

$(b)$ How many tickets should she buy to have a

$50%$ chance at winning?

Write fractions in the format “a/b”

$(a)$ (16) tickets

$(b)$ (8, 16/2) tickets

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Probability Formula

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

$(a)$ Find the number of tickets she should buy to have a certain chance at winning.

favourable outcomes

$=$ $x$ (number of tickets she should buy)

total outcomes

$=$ $16$ (the wheel has

$16$ sectors)

P(certain)

$=100%=1$

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

$1$	$=$	$\frac{\color{#e65021}{x}}{\color{#007DDC}{16}}$	Substitute values

$1$ $times16$	$=$	$x/16$ $times16$	Multiple both sides by $16$

$16$	$=$	$x$
$x$	$=$	$16$

$(b)$ Find the number of tickets she should buy to have a

$50%$ chance at winning.

favourable outcomes

$=$ $x$ (number of tickets she should buy)

total outcomes

$=$ $16$ (the wheel has

$16$ sectors)

P(50%)

$=1/2$

$\mathsf{P(}50\%\mathsf{)}$	$=$	$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$	Probability Formula

$1/2$	$=$	$\frac{\color{#e65021}{x}}{\color{#007DDC}{16}}$	Substitute values

$1/2$ $times16$	$=$	$x/16$ $times16$	Multiple both sides by $16$

$8$	$=$	$x$
$x$	$=$	$8$

$(a) 16$ tickets

$(b) 8$ tickets

Question 7 of 7

A spinning wheel has

$16$ sectors marked

$1$ -

$16$ . Each of these sectors corresponds to a ticket that can be bought and a lady named Jackie wants to buy some. Answer the following:

$(c)$ How many tickets should she buy to have a

$25%$ chance at winning?

$(d)$ What is the probability of her winning if she buys

$6$ tickets?

Write fractions in the format “a/b”

$(c)$ (4, 16/4) tickets

$(d)$ (⅜, 3/8, 6/16, 0.375, 0.38)

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Probability Formula

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

$(c)$ Find the number of tickets she should buy to have a

$25%$ chance at winning.

favourable outcomes

$=$ $x$ (number of tickets she should buy)

total outcomes

$=$ $16$ (the wheel has

$16$ sectors)

P(25%)

$=1/4$

$\mathsf{P(}25\%\mathsf{)}$	$=$	$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$	Probability Formula

$1/4$	$=$	$\frac{\color{#e65021}{x}}{\color{#007DDC}{16}}$	Substitute values

$1/4$ $times16$	$=$	$x/16$ $times16$	Multiple both sides by $16$

$4$	$=$	$x$
$x$	$=$	$4$

$(d)$ Find the probability of her winning if she buys

$6$ tickets.

favourable outcomes

$=$ $6$ (Jackie is about to buy

$6$ tickets)

total outcomes

$=$ $16$ (the wheel has

$16$ sectors)

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

	$=$	$\frac{\color{#e65021}{6}}{\color{#007DDC}{16}}$	Substitute values

	$=$	$\frac{3}{8}$

$(c) 4$ tickets

$(d) 3/8$

Simple Probability 1

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Quiz summary

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1. Question

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Well Done!

Range of Probability

2. Question

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Probability Formula

3. Question

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Probability Formula

4. Question

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Probability Formula

5. Question

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Probability Formula

6. Question

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Correct!

Probability Formula

7. Question

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Probability Formula

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