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Probability
How Likely?
In the real world events can not be predicted with total
certainty. The best we can do is say how likely they are to
happen, using the idea of probability.
Tossing a Coin
When a coin is tossed, there are two possible
outcomes:
•heads (H) or
•tails (T)
We say that the probability of the coin landing H is ½
Similarly, the probability of the coin landing T is 1/2.
Throwing Dice
When a single die is thrown, there are six possible outcomes:
1, 2, 3, 4, 5, 6
The probability of throwing any one of these numbers is 1/6.
Probability of an event happening
= Number of ways it can happen
Total number of outcomes
Example: the chances of rolling a "4" with a die
Number of ways it can happen: 1
(there is only 1 face with a "4" on it)
Total number of outcomes: 6 (there are 6 faces altogether)
So the probability = 1/6
Probability Line
You can show probability on a Probability Line:
The probability is always between 0 and 1
Probability does not tell us exactly what will happen, it is just a
guide.
Probability Definitions
A trial is a single act by which an observation is noted e.g. a roll
of a dice.
An experiment is the process by which an observation is noted
and consists of one or more trials.
When a statistical experiment is conducted, there are a number
of possible outcomes. The possible outcomes are called a
sample space and are often denoted by the letter S.
An event is a subset of a sample space.
The impossible event (or empty set) is one that contains no
outcomes. It is often denoted by the Greek letter Ф (phi).
Note
• The definition P(A) = Number of outcomes that yield event A
Total number of possible outcomes
makes sense only if the number of possible outcomes is finite.
We will only consider these cases.
• If an event can never occur, its probability is 0. An event that
always happens has a probability of 1
• The value of a probability must always lie between 0 and 1
- a result outside this range indicates an error has occurred
• A probability may be expressed as a decimal or fraction
• All outcomes are considered equally likely
Here we can record data and predict outcomes based on past data
– Tennis player A has beaten player B in 8 out of the last 10
matches – we could say there is an 80% chance of them winning
the next match
Subjective probabilities differ from person to person. Because the
probability is subjective, it contains a high degree of personal bias.
An example of subjective probability could be asking a sports
journalist , before the football season starts, the chances of St
George winning the grand final. While there is no absolute
mathematical proof behind the answer to the example, he might
still reply in actual percentage terms, such as they have a 25%
chance of winning the grand final – it is an educated guess, an
estimate without any real data to support it.
Probability of throwing a head or a tail is
Probability of a Head + probability of a tail = 1/2 +1/2 = 1
that is we can be absolutely certain you will throw a head or a tail
But what is the probability of a head and then a tail?
when you have the word "AND" - that is two events must occur it is
harder and we multiply the probabilities 1/2 x 1/2 = 1/4
Prove this using a Tree Diagram …..
One more example - the probability of finding someone who is 6 foot
four AND has blue eyes is much more remote than finding someone
who is 6 foot four OR has blue eyes. The less chance something has of
occurring the closer the probability is to zero
I have $100,000 to invest - project B costs $80,000, project A costs $90,000 - if I
run with Project A - it will exclude project B and vice versa
More probability examples can be reviewed in the following video check
this out if you would like further examples…..
https://www.khanacademy.org/math/probability/independent-dependentprobability/addition_rule_probability/v/addition-rule-for-probability
Max, Susan and Karen earn over $50,000.
What is the probability of selecting someone from the
group at random who is Female or earns more than
$50,000 ?
P(F) = 2 / 3 , P (>$50,000) = 3/3
2/3 + (OR) 3/3 = 5/3 – (the overlap) 2/3 (two out of the
three belong to both groups) = 3/3 or 1
That is there is a 100% chance of selecting someone at
random who is female or earns over $50,000
YOUTUBE INDEPENDENT
EVENTS
http://www.khanacademy.org/math/probability/v/independentevents-1
Is it possible to toss a coin and get 14 heads in a row?
If the last 40 tosses have been heads what is the
chance that the next toss will also be a head?
http://www.mathsonline.com.au/r.html?lesson=5218
Suppose you have a box with 3 blue marbles, 2 red marbles, and
4 yellow marbles. You are going to pull out one marble, record
its colour, put it back in the box and draw another marble. What
is the probability of pulling out a red marble followed by a blue
marble?
The multiplication rule says we need to find P(red) P(blue).
P(red) = 2/9
P(blue) = 3/9
P(red and then a blue) = (2/9) x (3/9) = 6/81 = 2/27
The events in this example were independent. Once the first
marble was pulled out and its colour recorded, it was returned
to the box. Therefore, the probability for the second marble was
not effected by what happened on the first marble.
Consider the same box of marbles as in the previous example.
However in this case, we are going to pull out the first marble,
leave it out, and then pull out another marble. What is the
probability of pulling out a red marble followed by a blue marble?
We can still use the multiplication rule which says we need to find
P(red) P(blue). But be aware that in this case when we go to pull
out the second marble, there will only be 8 marbles left in the bag.
P(red) = 2/9
P(blue) = 3/8
P(red and then blue) = (2/9) x (3/8) = 6/72 = 1/12
The events in this example were dependent. When the first marble
was pulled out and kept out, it effected the probability of the
second event. This is what is meant by dependent events.
YOUTUBE
DEPENDENT EVENTS
http://www.khanacademy.org/math/probability/v/probabilityof-dependent-events
Or just focus on this event 2/3 chance of a
number less than 4 given that we know it
is odd
YOUTUBE PROBABILITY
WITH CARDS
http://www.khanacademy.org/math/probability/v/probabilitywith-playing-cards-and-venn-diagrams
Is it correct to say the probability of them not moving is 0.3?
Draw a tree diagram for this - example next slide
6 possible combinations
are HHTT, HTHT, HTTH,
TTHH, THTH, THHT
H
T
H
H
T
T
H
H
T
H
H
T
H
T
T
H
T
H
H
T
H
T
T
T
H
T
H
T
H
T
Or using the multiplication rule ½ x ½ x ½ because the first has to be green AND the second
has to be green AND the third has to be green
This looks like the same question but it is slightly different…
G
G
G
R
R
G
R
G
R
G
R
G
R
R
Given that 2 green
counters have already
been drawn we are now
just looking at the
probability that the third
green counter will also be
green. Here we are only
looking at the top last
branch of the tree - the
options are one in two
GGG or GGR
A different way of solving the same question ….
½x½
prob of
first
green
AND 2nd
green
½x½x½
Prob 1st
AND 2nd
AND 3rd
Because they are
independent events we
don’t need to do this
we just say each event
is ½ of drawing a green
counter
70 / 120 = .58
32 / 102 = .31
A table is a better idea - less confusing
Rich
Banker
3
Non Banker 1
Total
4
Not Rich
2
4
6
For (a) just focus on the rich column 3 out of four chances you are a
banker given that you are rich
For (b) just focus on the banker row - 3 of the five bankers are rich - 3/5 =
6 /10 = .6
Combinations / Permutations
There are three runners in a race A, B and C. List all the possible opportunities
that first and second place can be awarded to each runner? This is a
permutation question where order does matter – A,B is not the same as B, A
that is A first B second is not the same as B first and A second
On the calculator - 3, 2nd F , nPr , 2 =6
Proof AB, BA, BC, CB, AC, CA
I need to select two names from a hat containing three names, how many
different combinations of two names could I select?
Now order doesn’t matter – B,A is the same as A,B
On the calculator - 3, 2nd F , nCr 2 =3
AB, BC, AC
This question doesn’t exist in Ed 3
3 defective parts,
3 acceptable parts
I randomly select 2 out of the six parts what are all
the different combinations of defective parts
Work down the column, to find combinations e.g.
D1,D2, down to D1,D3, Down to D1, A4 etc
Using the Financial Calculator to find the number of
combinations 6, 2nd F , nCr 2 = 15
Answer….
107 = 10 000 000 different numbers
If digits were not repeated I think it would be a
permutation question ..with the calculator… 10
2nd F, nPr , 7 = 604,800 ... Test this on the next slide
Any assumption should be able to be tested by using smaller,
simpler examples.
So imagine the question said the company had a 2 digit serial
number and that numbers could be repeated the answer would be
102 = 100 – examples
00,01,02,03,04,05,06,07,08,09,10,11,12,13,14,15,16,17,18,19,20,
21,22,23,……….96,97,98,99,100
Now check the permutation solution …….
with the calculator… 10 2nd F, nPr , 2 = 90 …… the reason for the
difference is that with permutations you are saying imagine you
have 10 numbers in a hat how many combinations of two
numbers can you draw from the hat. Now you can’t pull out
00,11,22,33,44,55,66,77,88,99 – because there is only one of
each number in the hat.
(a) 6 x 3 x 5 = 90
(b) 7 x 6 x 3 x 5 = 630
Using the Financial Calculator to find the number of combinations
52, 2nd F , nCr 2 = 1326
This is a permutation question ..with the calculator… 10 2nd F, nPr , 5 = 30,240
Here order does matter - this is a permutation question ..so with the calculator…
7 2nd F, nPr , 3 = 210
One last comment on Permutations and Combinations – there will always be more
permutations than combinations because when order matters that is AB is not BA so
with Permutations we have possibilities yet with combinations AB is treated the same as
BA we only have one combination.
So in the above example A is on the left, B is in the middle and C is on the right is one
permutation and is different than B on the left, A is in the middle and C on the right etc.
If the question said how many combinations of photos are possible if each photo has
three people in it and there are seven people to choose from then the answer would be
7 2nd F, nCr , 3 = 35
Check out the table on the next slide for all the possible combinations
Seven people A,B,C,D,E,F,G, Combinations of three people in a
photo
1
ABC
11
ADF
21
BDF
31
CFG
2
ABD
12
ADG
22
BDG
32
DEF
3
ABE
13
AEF
23
BEF
33
DEG
4
ABF
14
AEG
24
BEG
34
DFG
5
ABG
15
AFG
25
BFG
35
EFG
6
ACD
16
BCD
26
CDE
7
ACE
17
BCE
27
CDF
8
ACF
18
BCF
28
CDG
9
ACG
19
BCG
29
CEF
10
ADE
20
BDE
30
CEG
Each one of these photo combinations can be arranged 6 ways. 35 combinations x 6 =
210 permutations – for example the people in photo 1 can be arranged
ABC,ACB,BCA,BAC,CAB or CBA
You Tube Video – Conditional Probability Bayers Theorem
http://www.youtube.com/watch?v=Zxm4Xxvzohk
Suggested Questions from Textbook……
Select a range of questions from the Problems in this chapter – enough so that you feel
comfortable with this topic