GCSE Probability powerpoint includes tree, venn and two way tables

GCSE: Probability
Dr J Frost (jfrost@tiffin.kingston.sch.uk)
Last modified: 26th
April 2017

GCSE Specification
208. Write probabilities using fractions, percentages or decimals
209. Compare experimental data and theoretical probabilities. Compare
relative frequencies from samples of different sizes.
210. Find the probability of successive events, such as several throws of a
single die.
Identify different mutually exclusive outcomes and know that the sum of the
probabilities of all these outcomes is 1.
211. Estimate the number of times an event will occur, given the probability
and the number of trials.
212. List all outcomes for single events, and for two successive events,
systematically. Use and draw sample space diagrams
213. Understand conditional probabilities. Use a tree diagram to calculate
conditional probability.
214. Solve more complex problems involving combinations of outcomes.
215. Understand selection with or without replacement. Draw a probability
tree diagram based on given information.

Probability of winning the UK lottery:
1 in 14,000,000 ___1___
14000000
0.000000714 0.0000714%
Odds Form
Fractional Form
Decimal Form Percentage Form
Which is best in this case?
?
?
? ?
RECAP: How to write probabilities

RECAP: Combinatorics
Combinatorics is the ‘number of ways of arranging something’.
We could consider how many things could do in each ‘slot’, then multiply these numbers
together.
How many 5 letter English words could there theoretically be?
B I L B O
26 x 26 x 26 x 26 x 26 = 265
1
How many 5 letter English words with distinct letters could there be?
S M A U G
26 x 25 x 24 x 23 x 22 = 7893600
2
How many ways of arranging the letters in SHELF?
E L F H S
5 x 4 x 3 x 2 x 1 = 5! (“5 factorial”)
3
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?
?
e.g.

STARTER: Probability Puzzles
Recall that:
In pairs/groups or otherwise, work out the probability of the following:
If I toss a coin twice, I see a Heads and
a Tails (in either order).
If I toss a coin three times, I see a 2
Heads and 1 Tail.
In 3 throws of a coin, a Heads never
follows a Tails.
Throwing three square numbers on a
die in a row.
Seeing exactly two heads in four throws of a coin.
I randomly pick a number from 1 to 4, four times,
and the values form a ‘run’ of 1 to 4 in any order
(e.g. 1234, 4231, ...).
After shuffling a pack of cards, the cards in each
suit are all together.
I have a bag of different colours of marbles and of
each. What’s the probability that upon picking of
them, they’re all of different colours?
1
2
3
4
5
N
NN
NN
N
OMG
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1. We might just know! 2. We can do an experiment and count
outcomes
For a fair die, we know
that the probability of
each outcome is , by
definition of it being a
fair die.
This is known as a:
Theoretical Probability
When we know the underlying
probability of an event.
We could throw the dice 100 times for
example, and count how many times we see
each outcome.
Outcome 1 2 3 4 5 6
Count 27 13 10 30 15 5
This is known as an:
Experimental Probability
Also known as the relative frequency , it is
a probability based on observing counts.
R.F.
?
?
How can we find the probability of an event?
?

Question 1: If we flipped a (not necessarily fair) coin 10 times
and saw 6 Heads, then is the true probability of getting a
Head?
Question 2: What can we do to make the experimental
probability be as close as possible to the true (theoretical)
probability of Heads?
No. It might for example be a fair coin: If we throw a fair coin 10 times we
wouldn’t necessarily see 5 heads. In fact we could have seen 6 heads! So the
relative frequency/experimental probability only provides a “sensible guess” for
the true probability of Heads, based on what we’ve observed.
Flip the coin lots of times. I we threw a coin just twice for example and saw 0
Heads, it’s hard to know how unfair our coin is. But if we threw it say 1000 times
and saw 200 heads, then we’d have a much more accurate probability.
The law of large events states that as the number of trials becomes large, the
experimental probability becomes closer to the true probability.
?
?
Check your understanding

A spinner has the letters A, B
and C on it. I spin the spinner
50 times, and see A 12 times.
What is the experimental
probability for P(A)?
Answer:
The probability of getting a 6
on an unfair die is 0.3. I throw
the die 200 times. How many
sixes might you expect to get?
Answer:
times
?
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RECAP: Estimating counts and probabilities

Test Your Understanding
The table below shows the probabilities for spinning an A, B and C on a spinner. If I
spin the spinner 150 times, estimate the number of Cs I will see.
Outcome A B C
Probability 0.12 0.34
P(C) = 1 – 0.12 – 0.34 = 0.54
Estimate Cs seen = 0.54 x 150 = 81
?
A
B
C
Outcome A B C
Count 30 45 45
I spin another spinner 120 times and see the
following counts:
What is the relative frequency of B?
45/120 = 0.375
?
A
B
A
B
C

So far…
208. Write probabilities using fractions, percentages or decimals
209. Compare experimental data and theoretical probabilities. Compare
relative frequencies from samples of different sizes.
210. Find the probability of successive events, such as several throws of a
single die.
Identify different mutually exclusive outcomes and know that the sum of the
probabilities of all these outcomes is 1.
211. Estimate the number of times an event will occur, given the probability
and the number of trials.
212. List all outcomes for single events, and for two successive events,
systematically. Use and draw sample space diagrams
213. Understand conditional probabilities. Use a tree diagram to calculate
conditional probability.
214. Solve more complex problems involving combinations of outcomes.
215. Understand selection with or without replacement. Draw a probability
tree diagram based on given information.




RECAP: Events
Examples of events:
Throwing a 6, throwing an odd number, tossing a heads, a randomly chosen person
having a height above 1.5m.
 The sample space is the set of all outcomes.
 An event is
a description of one or more outcomes. It is a subset of the sample
space.
𝜉
𝐴 𝐵
40
20
30
10
From Grade 6 you should be familiar with
representing sets using a Venn Diagram.
?
?
𝑃 ( 𝑨)=
1
3
We often use capital letters
to represent an event, then
use to mean the
probability of it.
The sample space

RECAP: Events
𝜉
𝐴 𝐵
40
20
30
10
=
n(A) =
P(A) =
P(A n B) =
n(A n B)=
n(A u B) =

When a fair coin is thrown, what’s the probability of:
And when 3 fair coins are thrown:
p(1st
coin H and 2nd
coin H and 3rd
coin H) =
Therefore in this particular case we found the following
relationship between these probabilities:
P(event1 and event2 and event3)
= P(event1) x P(event2) x P(event3)
1
8
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?
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Independent Events !

!Mutually Exclusive Events
! Independent Events
If A and B are mutually exclusive events, they can’t happen
at the same time. Then:
P(A or B) = P(A) + P(B)
If A and B are independent events, then the outcome of
one doesn’t affect the other. Then:
P(A and B) = P(A) P(B)

1 2 3 4 5 6 7 8
P(num divisible by 2) =
P(num divisible by 4) =
P(num divisible by 2 and by 4) =
1
2
1
4
1
4
Why would it have been wrong to multiply the
probabilities?
?
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But be careful…

Relevant exam style question…
Dave and Bob both come into school by bus from Hounslow.
The probability that Dave is late to school is 0.7.
The probability that Bob is late to school is 0.4.
Sheila claims that the probability Dave is late to school and Bob is late to
school is
Sheila is wrong. Explain why this might be.
The events are likely to not be independent, therefore we can’t
multiply the probabilities.
The events are connected, e.g. if Dave is late, then Bob may be
on the same bus and therefore more likely to be late too.
?

Getting a 6 on a die and a T on a coin. 

+ ×
Hitting a bullseye or a triple 20.  
+ ×
Getting a HHT or a THT after three
throws of an unfair coin (presuming we’ve
already worked out P(HHT) and P(THT).
 
+ ×
Getting 3 on the first throw of a die
and a 4 on the second. 

+ ×
Bart’s favourite colour being red and
Pablo’s being blue.


+ ×
Shaan’s favourite colour being red or blue.  
+ ×
Add or multiply probabilities?

Event 1


No Yes
Throwing a heads
on the first flip.
 
No Yes
 
No Yes
Event 2
Throwing a heads
on the second flip.
It rains tomorrow. It rains the day
after.
That I will choose
maths at A Level.
That I will choose
Physics at A Level.
Have a garden
gnome.
Being called Bart.


No Yes
Independent?

The probability that Kyle picks his nose today is 0.9. The probability that he
independently eats cabbage in the canteen today is 0.3. What’s the probability that
Kyle picks his nose, but doesn’t eat cabbage?
I pick two cards from the following. What is the probability the first number is a 1
and the second number a 2?
I throw 100 dice and 50 coins. What’s the probability I get all sixes and all heads?
a
b
c
?
1 2 2 3
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Question: Given there’s 5 red balls and 2 blue balls. What’s the
probability that after two picks we have a red ball and a blue ball?
R
B
R
B
R
B
5
7
2
7
4
6
2
6
5
6
1
6
?
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Tree Diagrams
After first pick, there’s less
balls to choose from, so
probabilities change.
Bro Tip: Note that probabilities
generally go on the lines, and
events at the end.

Question: Give there’s 5 red balls and 2 blue balls. What’s the
probability that after two picks we have a red ball and a blue ball?
R
B
R
B
R
B
5
7
2
7
4
6
2
6
5
6
1
6
5
21
5
21
?
?
P(red and blue) =
10
21
?
We multiply across the matching
branches, then add these values.
Tree Diagrams

...with replacement:
The item is returned before another is chosen.
The probability of each event on each trial is
fixed.
...without replacement:
The item is not returned.
•Total balls decreases by 1 each time.
•Number of items of this type decreases by 1.
Summary
Note that if the question doesn’t specify which, e.g. “You pick two balls from a
bag”, then PRESUME WITHOUT REPLACEMENT.

5
8
×
5
8
=
25
64
(3
8
×
5
8)+(5
8
×
3
8
?
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Example (on your sheet)
3
8
3
8
5
8
5
8
3
8
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Algebraic Probability Questions
Probability both orange:
This question famously made
national news…
b) Solve to find the value
of .
‘Cross multiply’
(𝑛+9)(𝑛−10)=0
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From DrFrostMaths homework platform…
-1 -6
? ?

Question 1
1
5
×
1
5
=
1
25
(1
5
×
4
5 )+(4
5
×
1
5)=
8
25
1 −
8
25
=
17
25
?
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?

Question 2
0.9
0.9
0.1
0.1 0.9
0.1
0.92
=0.81
2×0.1×0.9=0.18
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Question 3
5
14
9
14
4
13
9
13
5
13
8
13
two consonants?
9
14
×
8
13
=
36
91
( 5
14
×
9
13)+( 9
14
×
5
13)=
45
91
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Question 4
3
10
7
10
2
9
3
9
7
9
6
9
( 3
10
×
7
9)+( 7
10
×
3
9 )=
7
30
1 −
7
30
=
23
30
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Question 5 – “The Birthday Paradox”
𝟑𝟔𝟒
𝟑𝟔𝟓
𝟑𝟔𝟑
𝟑𝟔𝟓
𝟑𝟔𝟓
𝟑𝟔𝟓
×
𝟑𝟔𝟒
𝟑𝟔𝟓
×
𝟑𝟔𝟑
𝟑𝟔𝟓
×…×
𝟑𝟑𝟔
𝟑𝟔𝟓
=𝟎.𝟐𝟗𝟑𝟕…
That’s surprisingly likely!
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Question 7 (Algebraic Trees)
𝑝
4
𝑝
4
𝑝
4
×
𝑝
4
=
1
16
?

Question 8
𝑏
10
𝑏−1
9
?

Question N1
[Maclaurin M68] I have 44 socks in my drawer, each either red or black. In the dark I
randomly pick two socks, and the probability that they do not match is . How many of
the 44 socks are red?
Suppose there are red socks. There are therefore grey socks.
?

[Maclaurin 2013 Q4] Two coins are biased in such a way that, when they are both tossed
once:
(i) the probability of getting two heads is the same as the probability of getting two tails;
(ii) the probability of getting one head and one tail is .
For each coin, what is the probability of getting a head?
Question N2
Let the probability of coin 1 being Heads be and the probability of coin 2 being Heads .
Then using (i):
Then using (ii):
(and thus the other coin would be or )
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1904
4495
?
Answer =
Doing without a tree: Listing outcomes
BGG:
GBG:
GGB:
It’s usually quicker to just list
the outcomes rather than
draw a tree.
? Working

I have a bag consisting of 6 red balls, 4 blue and 3 green. I take three balls out of
the bag at random. Find the probability that the balls are the same colour.
RRR:
GGG:
BBB:
What’s the probability they’re of different colours:
RGB:
Each of the orderings of RGB will have the same probability.
So
Q
N
?
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Probability Past Paper Questions
Provided on sheet.
Remember:
1. List the possible events that match.
2. Find the probability of each (by multiplying).
3. Add them together.

Past Paper Questions
2
42
16
42
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222
380
?

64
110
?
?

When a fair coin is thrown, what’s the probability of:
And when 3 fair coins are thrown:
p(1st
coin H and 2nd
coin H and 3rd
coin H) =
Therefore in this particular case we found the following
relationship between these probabilities:
P(event1 and event2 and event3)
= P(event1) x P(event2) x P(event3)
1
8
?
?
?
Independent Events

If two events A and B are dependent, the first event does affect
the probability of the second event.
To find the probability of B, give that A has happened, use the
rule:
When you deal with Venn diagrams, this rule can be written in
set language as
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Dependent Events

GCSE Probability powerpoint includes tree, venn and two way tables

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