Download Basic Probability

Introduction to Probability
Section 7.3, 7.4, 7.5
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PRESENTED BY
E. G. GASCON
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Introduction to Probability
Section 7.3
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VOCABULARY & EXAMPLES
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Sample Space
The set of ALL possible outcomes for an experiment
is the sample space. (This will make up the
denominator of a probability)
Probability range is 0 ≤ P(E) ≤ 1
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Event
An event is a subset of a sample space defined by the
problem (the numerator)
Complement of an Event
• Is the set of ALL outcomes in a sample space that
are not included in the event. Denoted by E’
Example: If the event is throwing an 6 on a dice, then
the complement is throwing 1,2,3,4,or 5.
So… P(less than 6) = 1 – P(6)
This is sometimes an
easier way to find a
difficult probability
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Conditional Probability
 The Probability of an event occurring, given that another event has already





occurred. Denoted by
P(B|A), read Probability of B given A.
Ex P( Jack of Diamonds)
P(Jack | Diamonds)
There 1 jack of diamonds
P(Jack of Diamonds) = 1/52
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Independent Events
 Two event are independent if the occurrence of one
of the events does NOT affect the probability of the
occurrence of the other
 Use Conditional probability to determine
 P(B|A) = P(B) or P(A|B) = P(A) then they are
independent
Example of Independent Events: Draw two cards WITH replacement.
If the first card is a heart, and one puts it back in the deck, then if the second card drawn is red. The probability
of drawing a red card does not change because a heart was drawn first .
Example of Dependent Events: Draw two cards WITHOUT replacement.
If the first card is a heart, then if the second card drawn is red. The probability of drawing a red card changes
because a heart was drawn first, therefore there is one less card in the deck, and that heart might have been red.
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Mutually Exclusive Events
Events E and F are mutually exclusive events if E
= , meaning that they have no elements in
common.
One of these statements will be true:
•Event A and B cannot occur at the same time
•Event A and B have zero outcome in common
•P(A AND B) = 0
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F
Basic Probability Principle
Let S be a sample space of equally likely
outcomes, and let event E be a subset of S.
Then the probability that event E occurs is
Event
n( E )
P( E ) 
n( S )
note: 0
P(E)
Sample space
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Introduction to Probability
Section 7.4
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VOCABULARY & EXAMPLES
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Intersection Rule for Probability
(Multiplication Rule)
 Two events A AND B occur in sequence
 P(A and B) = P(A) * P(B|A)
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 SPECIAL CASE: If the events are INDEPENDENT
 P(A and B) = P(A) * P(B)
Take an extra minute to study this slide
It is important when faced with an AND
problem to decide if the two events are
independent (slide # 6).
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How to Solve an AND problem
Identify the events A and B in sequence
2. Decide whether the events are independent or
dependent
3. Find P(A), P(B), or if necessary P(B|A)
4. Use the appropriate Intersection rule
(Multiplication rule)
1.
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Example AND Problem
 Two cards are selected at random “without replacement,”
find the probability that a King , AND then a Queen is
selected.
 The events are a King then a Queen (Dependent Events)
P(K) =
4
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P(Q |K) =
4
51
4
16
4


.006
* 51
2652
52
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Example Special Case
 A coin is tossed and a die is rolled Find the probability of
getting a head and then rolling a 6
 The events are a Heads then a “2” (Independent Events)
P(H) =
1
2
P(“2”) =
1
1
* 6
2
1

 .083
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1
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OR
 For Mathematical purposes OR means
 INCLUSIVE “OR”
 Three ways for event A
 A occurs and B does not
 B occurs and A does not
 A and B both occur
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OR B to occur
 Ex: In a group, the number wearing Red shirts OR
Green Pants will be :

# of Red shirts with not green pants + # of green pants with
not red shirts + number of Red shirts and Green pants.
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Union Rule for Probability (Addition Rule)
For ANY events E OR F from a sample space S
P(E  F) = P(E) + P(F) – P(E  F)
Special Case:
For mutually exclusive events E AND F from a sample
space S,
P(E  F) = P(E) + P(F) (as in examples of section 7.3)
Take an extra minute to study this slide
It is important when faced with an OR
problem to decide if the two events are
mutually exclusive. (slide # 7)
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How to Solve an OR problem
Identify the events A and B
2. Decide whether the events are mutually exclusive
3. Find P(A), P(B), and if necessary P(A and B)
4. Use the appropriate Union rule (ADDITION rule)
1.
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Example: Special Case
Probability of drawing a spade OR heart.
Probability of drawing a spade.
n( Spades )
13
P( Spade) 

n(Cards in Deck ) 52
Probability of drawing a Heart.
n( Heart )
13
P( Heart ) 

n(Cards in Deck ) 52
Probability of drawing a spade OR heart.
P( Spade OR Heart )
n( Spade)
n( Hearts)


n(Cards in Deck ) n(Cards in Deck )
13 13 26 1 Note: These are mutually Exclusive events.




52 52 52 2
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Example
Probability of drawing a red face card.
Probability of drawing a red card.
n(Re dCard )
26
P(Re dCard ) 

n(Cards in Deck ) 52
Probability of drawing a face card.
n( FaceCard )
12
P( FaceCard ) 

n(Cards in Deck ) 52
There are 26 red cards
There are 12 face cards
BUT, these two events are NOT mutually exclusive...
so… there is an extra step
Probability of drawing a red face card.
n(Re d | FaceCard ) 6
P(Re d | FaceCard ) 

n(Cards in Deck ) 52
There are 6 red –face cards
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Solution to Events NOT
mutually exclusive
Probability of drawing a red OR face card.
P(red OR FaceCard )
n(Re d )
n( FaceCard )
n(redFaceCards )



n(Cards in Deck ) n(Cards in Deck ) n(Cards in Deck )
26 12 6 32 8
    
52 52 52 52 13
Note: The difference between this problem and the example on slide 13
On the previous slide #13 was a problem that were mutually exclusive, and
this one they were not.
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Complement Rule
IF you know the probability that event E occurs, then
the probability that event E does NOT occur is:
P(E’) =1 − P(E)
Use this rule to save steps.
To find the probability of the roll of two fair dice yields a sum > 3, one could find
the
P(4) OR P(5) OR P(6) OR P(7) OR P(8) OR P(9) OR P(10) OR P(11) OR P(12)
BUT it is easier
Find the P( sum 3) = P(1) OR P(2) OR P(3) = 0 + (1/36) + (2/36) = (3/36) = 1/12
so…
P( sum >3) = 1 − P(sum  3) = 1 − (1/12) = 11/12
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Odds
The odds in favor of an event E are defined as the
ratio of P(E) to P(E’)
The odds that the roll of two fair dice yields a sum > 3 is…
P(sum >3) : P(sum  3) =
1
12  1:11
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Introduction to Probability
Section 7.5
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VOCABULARY & EXAMPLES
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Conditional Probability
Sometimes it is
not easy to
logically figure
the conditional
probability, so
we have a
formula
The conditional probability of event E given event F:
P  E | F
P(E  F)

P( F )
Also it can be stated that by re-writing the equation:
P(E  F)  P( F )  P  E | F 
or
P(E  F)  P( E )  P  F | E 
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Notice that
this formula
is a
derivation
from the
AND rule.
Example 7.5 # 17
The Problem
 Numbers 1,2,3,4,and 5 written on slips of paper, and
2 slips are drawn at random one at a time WITHOUT
replacement.
 Find the probability that the sum is 8, given the first
number is 5.
Sample space = {1,2,3,4,5}
P  sum 8 | 5
P("5"  "sum 8")

P("5")
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Solution
 Each number as a 1/5 chance of being drawn for the
first slip of paper .
 Each number that is left has a ¼ chance of being
drawn for the second slip of paper.
 There is only 1 way to get a “5” AND “sum of 8”,
(5 + 3).
 therefore…
1 1 1
P("5" AND " sum 8")  1   
 5 4  20
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Solutions cont.
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P (5) 
5
Putting it all together:
P  sum 8 | 5 
1
1 5 5 1
20


 

1 20 1 20 4
5
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Visual Solutions
Second Slip drawn
2
3 4
5
1
3
4
5
1 2
4
5
1
23
5
First Slip drawn
1
2
3
4
1
5
27
2
3
4
This is the only
scenario which
has a sum of 8,
and there is
only one way
to get it,
therefore the
probability of a
sum of 8 given
5 on the first
draw is 1/4
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