Joan Ridgway
All probabilities lie somewhere on a scale
between “Impossible” and “Certain”
0
All probabilities lie somewhere on a scale
between “Impossible” and “Certain”
The probability scale goes from
0 to 1
1
0
1
14,000,000
1
6
0.5
0.7
95%
Probabilities can be expressed either as fractions
or as decimals (and sometimes as percentages)
1
The probability of throwing a
6 with a fair dice is 1
P(6) =
6
1
6
So the probability of not
throwing a 6 is 5
6
P(not 6) = 1-
1
6
=
5
6
If the probability that it will
rain tomorrow is 0.7 …..
P(rain) = 0.7
Then the probability that it
will not rain tomorrow is 0.3
P(not rain) = 1 – 0.7 = 0.3
Suppose I toss a coin:
What is the probability of getting a head?
0.5, ½ or 50%
Suppose I toss two coins:
If I toss the coin twice, I would get one of these combinations:
Heads, Heads
Heads, Tails
Tails, Heads
Tails, Tails
or
H, H
H, T
T, H
T, T
What is the probability of getting two heads?
Only one of these four combinations is two heads
Suppose I toss two coins:
If I toss the coin twice, I would get one of these combinations:
Heads, Heads
Heads, Tails
Tails, Heads
Tails,Tails
or
H
H, H
H, T
T, H
T, T
What is the probability of getting two heads?
Only one of these four combinations is two heads
So the probability of getting a two heads in a row is
¼
A Sample Space is a list of all the possible outcomes,
e.g. HH, HT, TH, TT
We can show this in a Sample Space Diagram:
Second Coin
First Coin
H
T
H H, H
H, T
T
T, T
T, H
There are 4 possible outcomes if you toss a coin twice
So the probability of two heads is ¼
Suppose I throw a die.
There are 6 equally likely outcomes.
Suppose I throw two dice.
Suppose I throw two dice.
We can show the possible outcomes in a Sample Space Diagram:
Second Dice
First Dice
1
2
3
4
5
6
1
1, 1 1, 2 1, 3 1, 4 1, 5 1, 6
2
2, 1 2, 2 2, 3 2, 4 2, 5 2, 6
3
3, 1 3, 2 3, 3 3, 4 3, 5 3, 6
4
4, 1 4, 2 4, 3 4, 4 4, 5 4, 6
5
5, 1 5, 2 5, 3 5, 4 5, 5 5, 6
6
6, 1 6, 2 6, 3 6, 4 6, 5 6, 6
There are 36 (6 x 6) possible outcomes if you throw two dice.
If you throw two dice, what is the probability of getting a “double”?
Second Dice
First Dice
1
2
3
4
5
6
1
1, 1 1, 2 1, 3 1, 4 1, 5 1, 6
2
2, 1 2, 2 2, 3 2, 4 2, 5 2, 6
3
3, 1 3, 2 3, 3 3, 4 3, 5 3, 6
4
4, 1 4, 2 4, 3 4, 4 4, 5 4, 6
5
5, 1 5, 2 5, 3 5, 4 5, 5 5, 6
6
6, 1 6, 2 6, 3 6, 4 6, 5 6, 6
If you throw two dice, what is the probability of getting a “double”?
Second Dice
First Dice
1
2
3
4
5
6
1
1, 1 1, 2 1, 3 1, 4 1, 5 1, 6
2
2, 1 2, 2 2, 3 2, 4 2, 5 2, 6
3
3, 1 3, 2 3, 3 3, 4 3, 5 3, 6
4
4, 1 4, 2 4, 3 4, 4 4, 5 4, 6
5
5, 1 5, 2 5, 3 5, 4 5, 5 5, 6
6
6, 1 6, 2 6, 3 6, 4 6, 5 6, 6
6 out of the 36 possible outcomes are “doubles”,
6
so the probability is
36
If you throw two dice, what is the probability of getting a “double”?
Second Dice
First Dice
1
2
3
4
5
6
1
1, 1 1, 2 1, 3 1, 4 1, 5 1, 6
2
2, 1 2, 2 2, 3 2, 4 2, 5 2, 6
3
3, 1 3, 2 3, 3 3, 4 3, 5 3, 6
4
4, 1 4, 2 4, 3 4, 4 4, 5 4, 6
5
5, 1 5, 2 5, 3 5, 4 5, 5 5, 6
6
6, 1 6, 2 6, 3 6, 4 6, 5 6, 6
6 out of the 36 possible outcomes are “doubles”,
so the probability is 1
6
What is the probability of scoring 9 or more?
Second Dice
First Dice
1
2
3
4
5
6
1
1, 1 1, 2 1, 3 1, 4 1, 5 1, 6
2
2, 1 2, 2 2, 3 2, 4 2, 5 2, 6
3
3, 1 3, 2 3, 3 3, 4 3, 5 3, 6
4
4, 1 4, 2 4, 3 4, 4 4, 5 4, 6
5
5, 1 5, 2 5, 3 5, 4 5, 5 5, 6
6
6, 1 6, 2 6, 3 6, 4 6, 5 6, 6
What is the probability of scoring 9 or more?
Second Dice
First Dice
1
2
3
4
5
6
1
1, 1 1, 2 1, 3 1, 4 1, 5 1, 6
2
2, 1 2, 2 2, 3 2, 4 2, 5 2, 6
3
3, 1 3, 2 3, 3 3, 4 3, 5 3, 6
4
4, 1 4, 2 4, 3 4, 4 4, 5 4, 6
5
5, 1 5, 2 5, 3 5, 4 5, 5 5, 6
6
6, 1 6, 2 6, 3 6, 4 6, 5 6, 6
10 out of the 36 possible outcomes add up to 9 or
more, so the probability is 10
36
What is the probability of scoring 9 or more?
Second Dice
First Dice
1
2
3
4
5
6
1
1, 1 1, 2 1, 3 1, 4 1, 5 1, 6
2
2, 1 2, 2 2, 3 2, 4 2, 5 2, 6
3
3, 1 3, 2 3, 3 3, 4 3, 5 3, 6
4
4, 1 4, 2 4, 3 4, 4 4, 5 4, 6
5
5, 1 5, 2 5, 3 5, 4 5, 5 5, 6
6
6, 1 6, 2 6, 3 6, 4 6, 5 6, 6
10 out of the 36 possible outcomes add up to 9 or
5
more, so the probability is
18
We cannot always calculate the probability of an event;
sometimes we have to estimate it.
Suppose we did not know whether a dice was fair or weighted.
We could throw it 100 times to find how often we threw a 6.
We would expect to get a 6 about once in every 6 throws, as the
probability should be 1 in 6.
We cannot always calculate the probability of an event;
sometimes we have to estimate it.
Suppose we did not know whether a dice was fair or weighted.
We could throw it 100 times to find how often we threw a 6.
Tally
1
2
3
4
5
6
llll
llll
llll
llll
llll
llll
llll
llll
llll
llll
llll
llll
Frequency
llll
ll
lll
lll
ll
llll llll llll llll llll
15
12
13
13
12
35
100
We threw 35 sixes out of a total of 100 throws
We threw 35 sixes out of a total of 100 throws
The relative frequency of throwing a 6 with this dodgy dice is:
35
100
We threw 35 sixes out of a total of 100 throws
The relative frequency of throwing a 6 with this dodgy dice is:
35
100
or
0.35
The probability of throwing a 6 with a normal dice is:
1
6
or
.
0.16
We threw a 6 more than twice as often as would be expected.
The dice is probably weighted!!!
Go back to the coins.
Remember there were 4 possible outcomes if I toss 2 coins
Heads, Heads
Heads, Tails
Tails, Heads
Tails,Tails
or
H
H
T
T
H
T
H
T
There are 4 possible outcomes because 2 x 2 = 4, just as for
two dice there are 36 possible outcomes because 6 x 6 = 36
If I toss three coins, what are the possible combinations?
T T T
H
T
T
H
H
T
H
T
H
T
H
T
H
H
T
T
H
T
H
H
H
– 0 Heads (3 Tails)
1 Head (2 Tails)
2 Heads (1 Tail)
– 3 Heads
There are 8 possible
outcomes because
2x2x2=8
If I toss three coins, what are the possible combinations?
T T T
H
T
T
H
H
T
H
T
H
T
H
T
H
H
T
T
H
T
H
H
H
– 0 Heads (3 Tails)
The probability of 0 Heads is
1
8
1 Head (2 Tails)
The probability of 1 Head is
2 Heads (1 Tail)
3
8
3
8
– 3 Heads
1
The probability of 3 Heads is
8
The probability of 2 Heads is
If I toss three coins, what are the possible combinations?
H
H
H
T
H
T
T
T
H
H
T
H
T
H
T
T
H
T
H
H
T
T
H
T
Simulation Hyperlink
– 3 Heads
1
8
The probability of 3 Heads is
2 Heads (1 Tail)
3
8
The probability of 2 Heads is
1 Head (2 Tails)
The probability of 1 Head is
– 0 Heads (3 Tails)
The probability of 0 Heads is
3
8
1
8
Sometimes it is helpful to draw tree diagrams.
If I toss a coin twice, what is the probability of getting at least one head?
First coin
Second coin
1
.2
1
.2
1
.2
Heads
P(HH) =
1 1 1
 
2 2 4
Tails
P(HT) =
1 1 1
 
2 2 4
Heads
P(TH) =
1 1 1
 
2 2 4
Tails
P(TT) =
1 1 1
 
2 2 4
Heads
1
.2
1
.2
Tails
1
.2
Check that the probabilities add up to 1.
1 1 1 1
   1
4 4 4 4
The Watsons regard one boy and one girl as the ideal family. What is the
chance of getting one boy and one girl in their planned family of two?
First Child
Second Child
Combined
0.5
G
P(G,G) = 0.5 x 0.5 = 0.25
0.5
B
P(G,B) = 0.5 x 0.5 = 0.25
0.5
G
P(B,G) = 0.5 x 0.5 = 0.25
0.5
B
P(B,B) = 0.5 x 0.5 = 0.25
G
0.5
0.5
B
The probability of getting one of each is:
P(G,B) + P(B,G) = 0.25 + 0.25 = 0.5
Total = 1.00
Sometimes it is helpful to draw tree diagrams.
Joan travels to work on her bicycle. She has to go through two sets of traffic
lights on her way. At the first set of lights, the probability that they will be
green is 7/10. At the second set the probability that they will be green is 3/5
Sometimes it is helpful to draw tree diagrams.
Joan travels to work on her bicycle. She has to go through two sets of traffic
lights on her way. At the first set of lights, the probability that they will be
green is 7/10. At the second set the probability that they will be green is 3/5
1)
What is that probability that both sets will be green?
2)
What is the probability that she will have to stop once and only once at a
set of lights?
First set
Second set
3
.5
7
10
3
10
7 3 21
 
10 5 50
Green
P(both green) =
Not Green
P(green, not green) =
7 2 14
 
10 5 50
Green
P(not green, green) =
3 3 9
 
10 5 50
Not Green
P(both not green) =
3 2 6
 
10 5 50
Green
2
.5
3
.5
Not Green
2
.5
Check that the probabilities add up to 1.
21 14 9
6 50
  

1
50 50 50 50 50
Sometimes it is helpful to draw tree diagrams.
Joan travels to work on her bicycle. She has to go through two sets of traffic
lights on her way. At the first set of lights, the probability that they will be
green is 7/10. At the second set the probability that they will be green is 3/5
1)
What is that probability that both sets will be green?
2)
What is the probability that she will have to stop once and only once at a
set of lights?
First set
Second set
3
.5
7
10
3
10
P(both green) =
Not Green
P(green, not green) =
7
7 2 14

 
10 5 50 25
Green
P(not green, green) =
3 3 9
 
10 5 50
Not Green
P(both not green) =
3
3 2 6

 
10 5 50 25
Green
2
.5
3
.5
Not Green
2
.5
7 3 21
 
10 5 50
Green
You should simplify the probabilities where you can.
Sometimes it is helpful to draw tree diagrams.
Joan travels to work on her bicycle. She has to go through two sets of traffic
lights on her way. At the first set of lights, the probability that they will be
green is 7/10. At the second set the probability that they will be green is 3/5
1)
What is that probability that both sets will be green?
2)
What is the probability that she will have to stop once and only once at a
set of lights?
First set
Second set
3
.5
7
10
3
10
P(both green) =
Not Green
P(green, not green) =
7
7 2 14

 
10 5 50 25
Green
P(not green, green) =
3 3 9
 
10 5 50
Not Green
P(both not green) =
3
3 2 6

 
10 5 50 25
Green
2
.5
3
.5
Not Green
2
.5
1) P(both green) =
7 3 21
 
10 5 50
Green
21
50
Sometimes it is helpful to draw tree diagrams.
Joan travels to work on her bicycle. She has to go through two sets of traffic
lights on her way. At the first set of lights, the probability that they will be
green is 7/10. At the second set the probability that they will be green is 3/5
1)
What is that probability that both sets will be green?
2)
What is the probability that she will have to stop once and only once at a
set of lights?
First set
Second set
3
.5
7
10
3
10
P(both green) =
Not Green
P(green, not green) =
7
7 2 14

 
10 5 50 25
Green
P(not green, green) =
3 3 9
 
10 5 50
Not Green
P(both not green) =
3
3 2 6

 
10 5 50 25
Green
2
.5
3
.5
Not Green
2
.5
7 3 21
 
10 5 50
Green
2) P(only stopping once) =
14 9
23


50
50 50
Independent and Dependent Events
Two events are independent if they have no effect on each other.
If you choose two items with replacement the events will be independent
because the second choice will not be affected by the first choice.
There are 10 crayons in a bag; 5 blue ones, 3 red ones and 2 green
ones. I pick one at random, then put it back and choose another.
Draw a tree diagram to show the probabilities.
First Crayon
Second Crayon
5
10
Blue
Blue
3
10
2
10
5
10
3
10
Green
Blue
5
10
3
10
Red
2
10
2
10
Red
5
10
Green
2
10
Red
Green
Blue
3
10
Red
Green
What is the probability that both crayons will be green?
First Crayon
Second Crayon
5
10
Blue
Blue
3
10
2
10
5
10
3
10
Green
Blue
5
10
3
10
Red
2
10
2
10
Red
5
10
Green
2
10
Red
Green
Blue
3
10
Red
Green
2 2
4
 
10 10 100

1
25
What is the probability that I will choose one blue crayon
and one red crayon? (in either order)
First Crayon
Second Crayon
5
10
Blue
Blue
3
10
2
10
5
10
3
10
Green
or
Blue
5
10
3
10
Red
2
10
2
10
Red
5 3
15
 
10 10 100
5
10
Green
2
10
Red
Green
Blue
3
10
Red
Green
3 5
15
 
10 10 100
What is the probability that I will choose one blue crayon
and one red crayon? (in either order)
First Crayon
Second Crayon
5
10
Blue
Blue
3
10
2
10
5
10
3
10
Green
or
Blue
5
10
3
10
Red
2
10
2
10
Red
5 3
15
 
10 10 100
5
10
Green
2
10
Red
Green
Blue
3
10
3 5
15
 
10 10 100
Red
Green
15 15
30


100 100 100
3

10
Independent and Dependent Events
Two events are dependent if one event will affect the other. The
probability is said to be conditional.
If you choose two items without replacement the events will be dependent
because the first item you choose cannot be chosen again.
There are 10 jellybabies in a bag; 5 red ones, 3 green ones and 2 yellow
ones. I pick one at random and eat it, and then I choose another.
Draw a tree diagram to show the probabilities.
First Jellybaby
Second Jellybaby
Red
Red
Green
Yellow
5
10
3
10
Red
Green
Green
Yellow
Red
2
10
Yellow
Green
Yellow
First Jellybaby
Second Jellybaby
4
9
Red
2
9
5
10
3
10
Red
3
9
Green
Yellow
Red
Green
Green
Yellow
Red
2
10
Yellow
BE CAREFUL!!
Green
Yellow
If I choose a red one first
and eat it, there will be
only 9 jellybabies left, and
only 4 will be red ones!!
First Jellybaby
Second Jellybaby
4
9
Red
2
9
5
10
3
10
Red
3
9
Green
Yellow
Red
Green
Green
Yellow
Red
2
10
Yellow
Green
Yellow
First Jellybaby
Second Jellybaby
4
9
Red
Red
3
9
2
9
5
10
3
10
5
9
Green
2
9
2
10
Yellow
Green
BE CAREFUL!!
Yellow
Red
2
9
Green
Yellow
Red
Green
Yellow
If I choose a green one first
and eat it, there will be only
9 jellybabies left, and only 2
will be green ones!!
First Jellybaby
Second Jellybaby
4
9
Red
Red
3
9
2
9
5
10
3
10
5
9
Green
2
9
2
10
Yellow
Green
Yellow
Red
2
9
Green
Yellow
Red
Green
Yellow
First Jellybaby
Second Jellybaby
4
9
Red
Red
3
9
2
9
5
10
3
10
BE CAREFUL!!
Yellow
5
9
Red
2
9
Green
Green
2
9
2
10
Green
5
9
Yellow
1
9
Yellow
Red
3
9
Green
Yellow
If I choose a yellow one
first and eat it, there will be
only 9 jellybabies left, and
only 1 will be a yellow one!!
What is the probability that both jellybabies will be yellow?
First Jellybaby
Second Jellybaby
4
9
Red
Red
3
9
2
9
5
10
3
10
Yellow
5
9
Red
2
9
Green
Green
2
9
2
10
Green
5
9
Yellow
1
9
Yellow
Red
3
9
Green
Yellow
2 1 2
 
10 9 90

1
45
What is the probability that I will choose (and eat) one red
jelly baby and one green jelly baby? (in either order)
First Jellybaby
Second Jellybaby
4
9
Red
Red
3
9
2
9
5
10
3
10
5
9
Yellow
or
Red
2
9
Green
Green
2
9
2
10
Green
5 3 15
 
10 9 90
5
9
Yellow
1
9
Yellow
Red
3
9
Green
Yellow
3 5 15
 
10 9 90
What is the probability that I will choose (and eat) one red
jelly baby and one green jelly baby? (in either order)
First Jellybaby
Second Jellybaby
4
9
Red
Red
3
9
2
9
5
10
3
10
5
9
Yellow
or
Red
2
9
Green
5
9
Yellow
1
9
3 5 15
 
10 9 90
Green
2
9
2
10
Green
5 3 15
 
10 9 90
Yellow
Red
3
9
Green
Yellow
15 15 30


90 90 90
1

3