Understand the following
0  P( x)  1
P
(
x
)

1

number of favorable outcomes
Probability =
total possible outomes
2
Consider each of the following statements
Declare how you would best define the probability
that each will happen using one of the following
words
Unlikely
Impossible
Likely
Certain
Even Chance
The probability scale
Probability is a numerical measure of how likely or unlikely an event is to occur.
Probabilities are usually written as fractions, but can be written in any form
equivalent to that fraction. Eg ¾ = 0.75 = 75%
Probabilities can be anywhere between 0 (impossible) and 1 (certain):
Impossible
c
0
Unlikely
Even chance
b
Likely
d
Certain
a
½
1
a) an event with a probability of 0.8 would be described as very likely
b) an event with a probability of 0.4 would be described as
unlikely
c) an event with a probability of 1/20 would be described as
very unlikely
d) an event with a probability of 6/12 would be described as
even chance
The probability scale
1. Complete this probability scale using the key words given
Impossible
0
Unlikely
Even
chance
Likely
½
Certain
Key words:
Even chance
Likely
Impossible
Certain
Unlikely
1
2. Label the events described below on the probability scale:
a) The chance of getting an even number when rolling a dice
b) The chance of winning the National Lottery
c) The chance of rain in March
b
0
a
½
3.Describe an event that you think has a probability of:
a) 0.3
_________________________________________________
b) 1
_________________________________________________
c) 0.8
_________________________________________________
c
1
Probability of an event
1. Bob is picking randomly from a bag containing tiles numbered 1 to 10.
Write down the probability that the number he picks is:
a) 7
1
10
b) 4 or less
4
10
 2 5 c) Odd
5
10
 12
d) A multiple of 3
3
10
2. A survey is conducted of pupils’ favourite team:
Team
Spurs
Man Utd
Liverpool
Arsenal
Pupils
12
8
4
6
John picks a pupil at random to ask more questions.
Write down the probability that the pupil he picks supports:
a) Liverpool
4
30
 2 15
b) A London team
18
3
30  5
c) Not Liverpool
3. A bag contains 20 coloured balls, some red and some blue.
Keith knows that the probability of picking a red ball is 2/5.
How many red balls are there?
8 red balls
13
15
4
2
5

4
8
20
3. The table shows information about the number of goals scored by Aston Villa
in each game of the season so far:
Number
of goals
Number
of games
0
3
1
6
2
5
3
4
More
than 3
2
Mr Walker has all the games on DVD and decides to
watch one. He picks a game randomly.
What is the probability he picks a game with:
a) exactly 1 goal by Aston Villa,
6
3

20 10
b) 2 or more goals by Aston Villa
11
20
Total = 20 games
4. The cumulative frequency curve
below shows the distribution of the
height of 50 students. Estimate the
probability that a student picked at
random will be more than 164cm tall
10 1

50 5
Total probability of events
If one of a set of possible outcomes xi must happen, then ΣP(xi) = 1
The sum of their probabilities is 1
Eg the probability of rain today is 0.7
so the probability of no rain is 0.3
Eg a coin is biased so that the probability of heads is 3/5 so the probability of tails is 2/5
Eg a 4-sided spinner has the following probabilities of getting each number.
The probability that the spinner will land on 2 is equal to the probability it will
land on 4. Complete the table.
Number
1
2
3
4
0.2  x  0.3  x  1
Probability
0.2
x
0.3
x
 2 x  0.5  1
 2 x  0.5
 x  0.25
Total probability of events
1. A die is biased so that the probability of rolling a six is
What is the probability of not rolling a six?
1
4
.
1 41  34
2. A die is biased so that the probability of each number is:
Number
1
2
3
4
5
6
Probability
0.1
0.15
x
0.2
x
0.05
ΣP(xi) = 1
Find the value of x
0 .5  2 x  1
 x  0.25
3. The weatherman claims that it is twice as likely to snow as not.
Complete the table:
Snow
Probability
2
3
No snow
1
3
At the races
Each horse moves 1 square if you get its
total when you roll both dice. Is it a fair race?
2
3
4
5
6
7
8
9
10
11
12
Why isn’t the race fair?
Consider the possible outcomes from finding the total of 2 dice:
Dice A
Dice B
1
2
1
2
3
2
3
4
3
4
5
6
4
5
6
7
5
6
7
8
3
2
1
36
3
2
4
5
6
5
7
8
4
3
6
5
4
8
9
10
6
5
11
7
6
10 11 12
8
5
9
4
10
3
11
2
12
1
36
7
8
9
7
8
9
5
Probability
6
7
6
4
Total
9
10
Number of desired outcomes
Probabilit y 
Number of possible outcomes
7 is the most likely total, so horse 7 is most likely to win
2 & 12 are the least likely totals, so horses 2 and 12 are least likely to win
36
36
36
36
36
36
36
36
36
Probability using tables
Eg in a game, two fair dice are rolled and a score is found by multiplying the
numbers obtained together.
a) Show the possible outcomes in the table below
b) Use your completed table to find the probability of getting a score of 12
c) Use the table to find the probability of getting a score of 23 or more
Dice A
Dice B
a)
1
2
3
4
5
6
1
1
2
3
4
5
6
2
2
4
6
8
10
12
3
3
6
9
12
15
18
4
4
8
12
16
20
24
5
5
10
15
20
25
30
6
6
12
18
24
30
36
Probabilit y 
Number of desired outcomes
Number of possible outcomes
b) 4 outcomes out of 36
give a score of 12
Probability 
4
1

36 9
c) 6 outcomes out of 36
give a score of 23+
Probability 
6
1

36 6
Probability using tables
1. In a game, two fair spinners are spun and a score is found
by adding the numbers obtained together.
a) Show the possible outcomes in the table below
b) Use your table to find the probability of getting a score of 7
c) Find the probability of getting a score of 4 or less
Spinner A
Spinner B
a)
1
2
3
4
1
2
3
4
5
2
3
4
5
6
3
4
5
6
7
4
5
6
7
8
b) 2 outcomes out of 16
give a score of 7
Probability 
c) 6 outcomes out of 16
give a score of 4 or less
Probability 
Probabilit y 
2
1

16 8
Number of desired outcomes
Number of possible outcomes
6
3

16 8
2. Keith has 3 coloured balls in a bag- red, blue and yellow.
He picks one, records its colour, puts it back and picks another.
a) Complete the table to show the possible outcomes
b) Write down the probability Keith picks:
i) two balls of the same colour
bi) 3 outcomes out of 9
ii) two balls of different colour
give the same colour
iii) at least one yellow ball
1st pick
2nd pick
a)
R
B
Y
R
RR
BR
YR
B
RB
BB
YB
Y
Probabilit y 
RY
BY
YY
Number of desired outcomes
Number of possible outcomes
Probability 
3
1

9
3
ii) Either they are the
same colour or not
Probability  1 
1 2

3 3
iii) 5 outcomes out of 9
have a yellow ball
Probability 
5
9
Tree diagrams
Sometimes, a tree diagram can help you understand probabilities
Eg a coin is biased so that the probability of throwing heads each time is 2/3
The branches show the possible
outcomes and their probabilities
1st throw
2nd throw
2
2
Any chain of branches from the
beginning to the end represents
a combination of outcomes
3
H
Two heads in a row
1
3
T
Heads followed by tails
3
H
Tails followed by heads
1
3
T
Two tails in a row
H
3
1
3
2
T
Tree diagrams
Eg Johnny has a 0.4 chance of scoring from a free-kick and a 0.7 chance of
scoring from a penalty
The branches show the possible
outcomes and their probabilities
Free-kick
Penalty
0.7
0.4
0.6
Any chain of branches from the
beginning to the end represents
a combination of outcomes
Score
Scores both
Miss
Scores free-kick but misses penalty
Score
Misses free-kick but scores penalty
Miss
Misses both
Score
0.3
0.7
Miss
0.3
Finding probabilities with tree diagrams
Eg a coin is biased so that the probability of throwing heads each time is 2/3
To find the probability of a combination of outcomes,
multiply the probabilities along the relevant branches
1st throw
2nd throw
2
2
3
H
P(Two heads in a row)
1
3
T
P(Heads followed by tails)
 2 3  13  2 9
3
H
P(Tails followed by heads)

1
3
T
P(Two tails in a row)
H
2
1
3
 23  23  49
3
T
If more than one combination gives the
desired outcome, add their probabilities

P(one head, one tail)
1
3
1
3
 23  29
 13 
1
9
 29  29  49
Finding probabilities with tree diagrams
Eg Johnny has a 0.4 chance of scoring from a free-kick and a 0.7 chance of
scoring from a penalty
To find the probability of a combination of outcomes,
multiply the probabilities along the relevant branches
Freekick
Penalty
0.7
Score
0.3
Miss
P(scores both)
 0.4  0.7  0.28
P(scores one)
 0.4  0.3  0.6  0.7
Score
0.4
0.6
0.7
Score
0.3
Miss
 0.54
Miss
P(scores neither)
 0.6  0.3  0.18
If more than one combination gives the desired outcome, add their probabilities
Tree diagrams
1. Simon plays one game of tennis and one game of snooker.
The probability that Simon will win at snooker is
The probability that Simon will win at tennis is
3
4
1
3
tennis
a) Complete the tree diagram
b) Work out the probability that
Simon wins both games.
3 1 1
 
4 3 4
c) Work out the probability that
Simon will win only one game.
3 2 1 1 6
1
7
   


4 3 4 3 12 12 12
3
4
1
4
..........
snooker
1
3
Simon
wins
2
..........
3
Simon
does not
win
1
..........
3
Simon
wins
2
..........
3
Simon
does not
win
Simon
wins
Simon
does not
win
2. Julie and Pat are going to the cinema.
The probability that Julie will arrive late is 0.2
The probability that Pat will arrive late is 0.6
The two events are independent.
Pat
a) Complete the diagram.
late
0.6
Julie
late
0.4
0.2
not
late
b) Work out the probability that
Julie and Pat will both arrive late.
0.2  0.6  0.12
c) Work out the probability that
neither of them arrive late.
0.8  0.4  0.32
0.8
late
0.6
not
late
0.4
not
late
3. Julie throws a fair red dice once and a fair blue dice once.
a) Complete the probability tree diagram to show the outcomes.
Label clearly the branches of the probability tree diagram.
The probability tree diagram has been started in the space below.
Red
Dice
b) Calculate the probability that
Julie gets at least one six.
= 1 – P(no sixes)
5 5
 1 
6 6
 1
25 11

36 36
Blue
Dice
1
6
1
6
5
6
Six
Six
Not
Six
5
6 1
6
5
6
Not
Six
Six
Not
Six
4. Loren has two bags.
The first bag contains 3 red counters and 2 blue counters.
The second bag contains 2 red counters and 5 blue counters.
Loren takes one counter at random from each bag.
Counter from
first bag
a) Complete the probability tree diagram.
Counter from
second bag
Red
2
7
b) Work out the probability that Loren
takes one counter of each colour.
Red
3 5 2 2 15 4
19
   


5 7 5 7 35 35 35
3
5
5
......
7
2
......
7
2
......
5
Blue
Red
Blue
5
......
7
Blue
1
2
3
4
5
Play your cards right
6
7
8
9
10
Play your cards right
1 2 3 4 5 6 7 8 9 10
P(Higher) 
6
2
4
1
7
8
4
9
7
8
5
7
6
6
3
5
2
4
Higher or lower???
The probability of higher or lower is conditional
on the cards that have already appeared
9
Non-replacement
Eg a bag contains 3 red and 7 blue balls.
A ball is picked, not replaced, and another picked.
Complete the tree diagram
There are 10 balls to choose
from when picking the 1st ball
1st pick
2nd pick
2
3
7
9
Red
If a red ball was picked first,
there are only 2 red balls left
Blue
If a red ball was picked first,
there are still 7 blue balls left
Red
If a blue ball was picked first,
there are still 3 red balls left
Blue
If a blue ball was picked first,
there are only 6 blue balls left
Red
10
7
3
10
9
9
Blue
6
9
If the object is not replaced, this affects
the probabilities on the 2nd pick
Non-replacement
Eg a bag contains 3 red and 7 blue balls.
A ball is picked, not replaced, and another picked.
Find the probability that:
To find the probability of a combination
a) 2 red balls are picked
of outcomes, multiply the probabilities
b) 1 of each colour is picked
along the relevant branches
1st pick
2nd pick
2
3
7
9
P(both red)
 3 10  2 9  6 90 
Red
Red
10
7
3
10
9
9
1
15
Blue
P(one of each)
 3 10  7 9  7 10  3 9
 2190  2190
Red
 42 90  7 15
Blue
6
9
Blue
If more than one combination gives the
desired outcome, add their probabilities
Non-replacement
1. A bag of sweets contains 2 toffees and 5 chocolates.
A sweet is picked, eaten, and another picked.
a) Complete the tree diagram
b) Find the probability that:
i) Both sweets are toffees
ii) 1 of each sweet is picked
2nd pick
1st pick
2
1
6
Toffee
5
Chocolate
bi) P(both toffee)
 2 7  16  2 42 
1
Toffee
7
6
bii) P(one of each)
 27  56  57  26
5
2
7
6
Toffee
Chocolate
4
6
Chocolate
 10 42  10 42
 10 21
21
2. 5 white socks and 3 black socks are in a drawer.
Stefan takes out two socks at random.
Work out the probability that Stefan takes out two socks of the same colour.
2nd pick
1st pick
4
5
7
White
P(both white)
 5 8  4 7  20 56
Black
P(both black)
 3 8  2 7  6 56
White
8
3
7
P(same colour)
3
5
8
7
White
Black
2
7
Black
 20 56  6 56
 26 56
 13 28
Listing outcomes systematically
Eg A bag contains 3 blue beads, 5 yellow beads and 2 green beads.
Sid takes a bead at random from the bag, records its colour and replaces it.
He does this two more times.
Work out the probability that, of the three beads Sid takes, exactly two are blue.
A tree diagram would take too long here
Combinations with
exactly 2 blue:
BBY
P(BBY) 
and P(BBY) = P(BYB) = P(YBB)
BYB
YBB
3
3
5
45



10 10 10 1000
P(BBG) 
3
3
2
18



10 10 10 1000
BBG
BGB
GBB
and P(BBG) = P(BGB) = P(GBB)
So P(exactly 2 blue) 
45
18
189
3 
3 
1000
1000
1000
Listing outcomes systematically
1. For any match, the probabilities of each result for Aston Villa are as follows:
1
P(win) =
2
1
P(draw) =
3
1
P(lose) =
6
Find the probability that, in 3 matches, Aston Villa win exactly 2 matches
Combinations with
exactly 2 wins:
WWD
P(WWD) 
and P(WWD) = P(WDW) = P(DWW)
WDW
DWW
1 1 1
1
  
2 2 3
12
P(WWL) 
1 1 1
1
  
2 2 6 24
WWL
WLW
LWW
and P(WWL) = P(WLW) = P(LWW)
So P(exactly 2 wins) 
1
1
3
3 
3 
12
24
8
2. A bag contains 2 blue balls and 3 green balls.
Pete takes a ball at random from the bag, records its colour and replaces it.
He does this two more times.
Work out the probability that, of the three balls Pete takes, exactly two are the
same colour. (Hint – what is the alternative to 2 being the same colour?)
‘2 the same colour’ means 1 is a different colour
The only other option is all 3 are the same colour
So P(2 the same colour) = 1 – P(all the same colour)
3
3
8
27
7
35
 2  3 
P(all the same colour) = P(BBB) + P(GGG)       



 5   5  125 125 125 25
So P(2 the same colour) = 1
7
18

25
25
Expectation
Expectation is the long-run average you would get if a test was repeated many times
If an event has probability p, the expectation in n trials is np
Expectation is used as an estimate for how many times an event will occur
Eg a coin is biased so that the probability of throwing heads is ¾.
Dave is going to throw the coin 200 times.
Work out an estimate for the number of times the coin lands on heads.
n = 200 and p = ¾
so expectation = np = 200 x ¾ = 150
Eg There are 306 MPs in the Conservative Party.
4/ of them say they support proposals to increase tuition fees.
5
Work out an estimate for the number who will vote in favour of the changes
n = 306 and p = 4/5 so expectation = np = 306 x 4/5 = 244.8 = 245 to nearest integer
Expectation
1. A coin is biased so that the probability of getting heads is 3/5.
Dave is going to throw the coin 120 times.
Work out an estimate for the number of times the coin lands on tails.
Expectation = np = 120 x 2/5 = 48
2. The chance of rain each day in April is 2/3.
Estimate the number of days you can expect rain in April.
30 days in April
Expectation = np = 30 x 2/3 = 20
3. A door-to-door salesman achieves sales with a probability of 3/10.
How many doors must he approach in order to expect an average of 15 sales a day?
If 3/10 x n = 15 then n = 15  310 = 50 doors
Expectated winnings
Keith designs a game. It costs £1.60 to play the game.
The probability of winning the game is 2/5
The prize for each win is £3
80 people play the game.
Work out an estimate of the profit that Keith should expect to make.
Takings = 80 x 1.6 = £128
Expected winners = 2/5 x 80 = 32
Expected payout = 32 x 3 = £96
Estimated profit = 128 - 96 = £32
Profit = takings - costs
Expectation = np
4. A fruit machine costs £1 to play and pays out £40 with a probability of 1/20.
Is the machine worth playing? Explain your answer.
Expected winnings each game = £40 x 1/20 = £2
But cost of game is only £1, so you can expect to win money in the long run
5. John and Tom play darts and pool every Saturday.
John wins at darts 2/5 of the time and wins at pool ¾ of the time.
a) Find the probability they win one of the games each.
Estimate the number of times they win one of the games each, over a 60 week period
Winner
at pool
Winner
at darts
3
2
5
b)
1
5
Tom
4
4
Tom
1
4
2 1 3 3 2
9
11
   


5 4 5 4 20 20 20
John
John
3
3
4
a)
John
Tom
Expectation = np
 60 
11
20
= 33 times
Experimental probability
Sometimes the probability of an event occuring is not understood (eg trying to
predict the stock market!) very well. Experimental data can be collected to give an
estimate of the actual probability.
If an event occurs x times in n trials, the probability of the event is approximated by x/n
Eg Bob is convinced his toast always lands butter side down when he drops it.
He drops a piece of toast and it lands butter side down 30 times in 50 attempts.
Comment on Bob’s claim.
x = 30 and n =50
so probability
 30 50  3 5
Bob’s claim is supported by the data,
although he has not conducted that many
trials so it is possible he was just unlucky.
The more trials, the more likely it is that the experimental data matches the actual theory
Eg Bob repeats the experiment, dropping the piece of toast 1000 times.
It lands butter side down 600 times. Comment on Bob’s claim now.
x = 600 and n =1000
so probability
 600 1000  3 5
Bob’s claim is more strongly supported
by the data, as it very unlikely he
would be ‘unlucky’ that many times.
Problem solving
Eg A bag contains some red counters and blue counters.
There are n red counters. There is 1 more blue counter than red counters.
Bob will take a counter at random from the bag, record the colour and pick another.
The probability that Bob picks two red counters is 1/6.
Prove that n  4
red
+ blue = total
n  (n  1)  2n  1
P(1st
n
pick red) =
2n  1
P(2nd pick red) =
n
n 1
n 1
n(n  1)



So P(both red) =
2n  1 2n
2n(2n  1) 4 n  2
n 1
1
1

,
giving
But P(both red) 
4n  2 6
6
 6(n  1)  4n  2
 6n  6  4n  2
 2n  8
n4
n 1
2n
Problem solving
1. A bag contains some black counters and white counters.
There are n black counters. There are 2 less white counters than black counters.
Bob will take a counter at random from the bag, record the colour and replaces it
before picking another.
The probability that Bob picks one of each counter is 3/8.
Find how many of each coloured counter there are in the bag
red
But P(one each) 
+ blue = total
n  (n  2)  2n  2
P(BW) =
P(WB) =
3
8
 16n(n  2)  3(2n  2) 2
n(n  2)
n
n2


2n  2 2n  2
(2n  2) 2
 16n 2  32n  12n 2  24n  12
n(n  2)
n2
n


2n  2 2n  2
(2n  2) 2
 n 2  2n  3  0
So P(one each) 
2n(n  2)
(2n  2) 2
 4n 2  8n  12  0
 (n  3)(n  1)  0
 n  3 as n > 0
So 3 black, 1 white counter in bag
2. Gary plays two games of chess against Mijan.
The probability that Gary will win any game against Mijan is 0.55
The probability that Gary will draw any game against Mijan is 0.3
In a game of chess, you score 1 point for a win, ½ point for a draw, 0 points for a loss.
Work out the probability that after two games, Gary’s total score will be the same as
Mijan’s total score.
P(Gary loses) = 0.15
Total scores are the same if:
Gary wins 1st game, Mijan wins 2nd
0.55  0.15  0.0825
Mijan wins 1st game, Gary wins 2nd
0.15  0.55  0.0825
Both games drawn
0.3  0.3  0.09
0.255

The types of people watching a film at a cinema are shown in the table.
TOTAL = 50
Two of these people are chosen at random to receive free cinema tickets.
Calculate the probability that the two people are adults of the same gender.
P(Adult Male, Adult Male) + P(Adult Female, Adult Female)
21
20
x
50
49
6
35
+
+
43
175
14
50
x
13
175
13
49
(4 marks)
Misconceptions
Discuss why each statement is incorrect
If you toss a fair coin and get heads 5 times in a
row, you are more likely to get tails the next time.
The probability is the same each
time- previous results are irrelevant
In a football match, you can either win, lose or
draw. So the probability of winning is 1/3.
Winning may not have the
same probability as losing
You are less likely to win with lottery numbers
1,2,3,4,5,6 than if you pick numbers at random
If you toss a coin 50 times and get
heads 40 times, the coin must be biased
Every number has the same chance
and so does every combination
It might be biased, as you would only
expect 25 heads, but it is still possible to
get 40 out of 50 heads with a fair coin.
If you roll two dice and add the results, the probability
of getting 9 is 1/11 as there are 11 possibilities (2-12)
When tossing a coin, you are just as likely to get
5 heads in a row as 10 in a row- it’s just chance
There are more ways to get
some totals than others
P(5 heads in a row) = 1/32
P(10 heads in a row) = 1/1024