MATH 110 Sec 13-2 Lecture: Operations With Events
Sometimes in mathematics, it is easier to solve a problem if
you can restate it in a different way.
MATH 110 Sec 13-2 Lecture: Operations With Events
Sometimes in mathematics, it is easier to solve a problem if
you can restate it in a different way.
For example, if you are calculating 𝑃(𝐸), the probability of some
event 𝐸, but the calculation is complicated, sometimes it is easier
to instead calculate 𝑃(𝐸 ′ ) , the probability of the complement of
event 𝐸. This is useful because of the result below.
MATH 110 Sec 13-2 Lecture: Operations With Events
Sometimes in mathematics, it is easier to solve a problem if
you can restate it in a different way.
For example, if you are calculating 𝑃(𝐸), the probability of some
event 𝐸, but the calculation is complicated, sometimes it is easier
to instead calculate 𝑃(𝐸 ′ ) , the probability of the complement of
event 𝐸. This is useful because of the result below.
PROBABILITY OF THE
COMPLEMENT OF AN EVENT
If 𝐸 is an event, then 𝑃 𝐸 = 1 − 𝑃(𝐸 ′ ).
MATH 110 Sec 13-2 Lecture: Operations With Events
Sometimes in mathematics, it is easier to solve a problem if
you can restate it in a different way.
For example, if you are calculating 𝑃(𝐸), the probability of some
event 𝐸, but the calculation is complicated, sometimes it is easier
to instead calculate 𝑃(𝐸 ′ ) , the probability of the complement of
event 𝐸. This is useful because of the result below.
PROBABILITY OF THE
COMPLEMENT OF AN EVENT
If 𝐸 is an event, then 𝑃 𝐸 = 1 − 𝑃(𝐸 ′ ).
MATH 110 Sec 13-2 Lecture: Operations With Events
For example, given the chart below about a group of voters’ party
affiliation, if the event 𝐸 is ‘Person has a party affiliation’, find P(𝐸).
MATH 110 Sec 13-2 Lecture: Operations With Events
For example, given the chart below about a group of voters’ party
affiliation, if the event 𝐸 is ‘Person has a party affiliation’, find P(𝐸).
Because we are dealing with probabilities, we first need to convert the
percents to decimals. (Remember, the probabilities must sum to 1.)
MATH 110 Sec 13-2 Lecture: Operations With Events
For example, given the chart below about a group of voters’ party
affiliation, if the event 𝐸 is ‘Person has a party affiliation’, find P(𝐸).
Because we are dealing with probabilities, we first need to convert the
percents to decimals. (Remember, the probabilities must sum to 1.)
%
=0.237
7.2%=0.072
3.5%=0.035
4.4%=0.044
%
=0.371
%
=0.241
MATH 110 Sec 13-2 Lecture: Operations With Events
For example, given the chart below about a group of voters’ party
affiliation, if the event 𝐸 is ‘Person has a party affiliation’, find P(𝐸).
The answer is
𝑃 ℎ𝑎𝑠 𝑎𝑓𝑓𝑖𝑙𝑖𝑎𝑡𝑖𝑜𝑛 = 𝑃 𝐷𝑒𝑚 + 𝑃 𝑅𝑒𝑝 + 𝑃 𝐺𝑟 + 𝑃 𝐿𝑖𝑏 + 𝑃 𝑂𝑡ℎ𝑒𝑟
%
=0.237
7.2%=0.072
3.5%=0.035
4.4%=0.044
%
=0.371
%
=0.241
MATH 110 Sec 13-2 Lecture: Operations With Events
For example, given the chart below about a group of voters’ party
affiliation, if the event 𝐸 is ‘Person has a party affiliation’, find P(𝐸).
The answer is
𝑃 ℎ𝑎𝑠 𝑎𝑓𝑓𝑖𝑙𝑖𝑎𝑡𝑖𝑜𝑛 = 𝑃 𝐷𝑒𝑚 + 𝑃 𝑅𝑒𝑝 + 𝑃 𝐺𝑟 + 𝑃 𝐿𝑖𝑏 + 𝑃 𝑂𝑡ℎ𝑒𝑟
𝑃 ℎ𝑎𝑠 𝑎𝑓𝑓𝑖𝑙𝑖𝑎𝑡𝑖𝑜𝑛 = 0.371 + 0.241 + 0.044 + 0.035 + 0.072 = 0.763
%
=0.237
7.2%=0.072
3.5%=0.035
4.4%=0.044
%
=0.371
%
=0.241
MATH 110 Sec 13-2 Lecture: Operations With Events
For example, given the chart below about a group of voters’ party
affiliation, if the event 𝐸 is ‘Person has a party affiliation’, find P(𝐸).
The answer is
𝑃 ℎ𝑎𝑠 𝑎𝑓𝑓𝑖𝑙𝑖𝑎𝑡𝑖𝑜𝑛 = 𝑃 𝐷𝑒𝑚 + 𝑃 𝑅𝑒𝑝 + 𝑃 𝐺𝑟 + 𝑃 𝐿𝑖𝑏 + 𝑃 𝑂𝑡ℎ𝑒𝑟
𝑃 ℎ𝑎𝑠 𝑎𝑓𝑓𝑖𝑙𝑖𝑎𝑡𝑖𝑜𝑛 = 0.371 + 0.241 + 0.044 + 0.035 + 0.072 = 0.763
%
=0.237
7.2%=0.072
3.5%=0.035
4.4%=0.044
%
=0.371
%
=0.241
MATH 110 Sec 13-2 Lecture: Operations With Events
For example, given the chart below about a group of voters’ party
affiliation, if the event 𝐸 is ‘Person has a party affiliation’, find P(𝐸).
The answer is
𝑃 ℎ𝑎𝑠 𝑎𝑓𝑓𝑖𝑙𝑖𝑎𝑡𝑖𝑜𝑛 = 𝑃 𝐷𝑒𝑚 + 𝑃 𝑅𝑒𝑝 + 𝑃 𝐺𝑟 + 𝑃 𝐿𝑖𝑏 + 𝑃 𝑂𝑡ℎ𝑒𝑟
𝑃 ℎ𝑎𝑠 𝑎𝑓𝑓𝑖𝑙𝑖𝑎𝑡𝑖𝑜𝑛 = 0.371 + 0.241 + 0.044 + 0.035 + 0.072 = 0.763
But it’s easier to use the previous result:
%
=0.237
7.2%=0.072
3.5%=0.035
4.4%=0.044
%
=0.371
%
=0.241
MATH 110 Sec 13-2 Lecture: Operations With Events
For example, given the chart below about a group of voters’ party
affiliation, if the event 𝐸 is ‘Person has a party affiliation’, find P(𝐸).
The answer is
𝑃 ℎ𝑎𝑠 𝑎𝑓𝑓𝑖𝑙𝑖𝑎𝑡𝑖𝑜𝑛 = 𝑃 𝐷𝑒𝑚 + 𝑃 𝑅𝑒𝑝 + 𝑃 𝐺𝑟 + 𝑃 𝐿𝑖𝑏 + 𝑃 𝑂𝑡ℎ𝑒𝑟
𝑃 ℎ𝑎𝑠 𝑎𝑓𝑓𝑖𝑙𝑖𝑎𝑡𝑖𝑜𝑛 = 0.371 + 0.241 + 0.044 + 0.035 + 0.072 = 0.763
But it’s easier to use the previous result:
If 𝐸 is an event, then 𝑃 𝐸 = 1 − 𝑃(𝐸 ′ ).
%
=0.237
7.2%=0.072
3.5%=0.035
4.4%=0.044
%
=0.371
%
=0.241
MATH 110 Sec 13-2 Lecture: Operations With Events
For example, given the chart below about a group of voters’ party
affiliation, if the event 𝐸 is ‘Person has a party affiliation’, find P(𝐸).
The answer is
𝑃 ℎ𝑎𝑠 𝑎𝑓𝑓𝑖𝑙𝑖𝑎𝑡𝑖𝑜𝑛 = 𝑃 𝐷𝑒𝑚 + 𝑃 𝑅𝑒𝑝 + 𝑃 𝐺𝑟 + 𝑃 𝐿𝑖𝑏 + 𝑃 𝑂𝑡ℎ𝑒𝑟
𝑃 ℎ𝑎𝑠 𝑎𝑓𝑓𝑖𝑙𝑖𝑎𝑡𝑖𝑜𝑛 = 0.371 + 0.241 + 0.044 + 0.035 + 0.072 = 0.763
But it’s easier to use the previous result:
If 𝐸 is an event, then 𝑃 𝐸 = 1 − 𝑃(𝐸 ′ ).
%
=0.237
%
The complement of
=0.371
‘has party affiliation’ is
7.2%=0.072
‘has no party affiliation’.
3.5%=0.035
4.4%=0.044
%
=0.241
MATH 110 Sec 13-2 Lecture: Operations With Events
For example, given the chart below about a group of voters’ party
affiliation, if the event 𝐸 is ‘Person has a party affiliation’, find P(𝐸).
The answer is
𝑃 ℎ𝑎𝑠 𝑎𝑓𝑓𝑖𝑙𝑖𝑎𝑡𝑖𝑜𝑛 = 𝑃 𝐷𝑒𝑚 + 𝑃 𝑅𝑒𝑝 + 𝑃 𝐺𝑟 + 𝑃 𝐿𝑖𝑏 + 𝑃 𝑂𝑡ℎ𝑒𝑟
𝑃 ℎ𝑎𝑠 𝑎𝑓𝑓𝑖𝑙𝑖𝑎𝑡𝑖𝑜𝑛 = 0.371 + 0.241 + 0.044 + 0.035 + 0.072 = 0.763
But it’s easier to use the previous result:
If 𝐸 is an event, then 𝑃 𝐸 = 1 − 𝑃(𝐸 ′ ).
%
=0.237
%
The complement of
=0.371
‘has party affiliation’ is
7.2%=0.072
‘has no party affiliation’.
3.5%=0.035
4.4%=0.044
%
=0.241
MATH 110 Sec 13-2 Lecture: Operations With Events
For example, given the chart below about a group of voters’ party
affiliation, if the event 𝐸 is ‘Person has a party affiliation’, find P(𝐸).
The answer is
𝑃 ℎ𝑎𝑠 𝑎𝑓𝑓𝑖𝑙𝑖𝑎𝑡𝑖𝑜𝑛 = 𝑃 𝐷𝑒𝑚 + 𝑃 𝑅𝑒𝑝 + 𝑃 𝐺𝑟 + 𝑃 𝐿𝑖𝑏 + 𝑃 𝑂𝑡ℎ𝑒𝑟
𝑃 ℎ𝑎𝑠 𝑎𝑓𝑓𝑖𝑙𝑖𝑎𝑡𝑖𝑜𝑛 = 0.371 + 0.241 + 0.044 + 0.035 + 0.072 = 0.763
But it’s easier to use the previous result:
If 𝐸 is an event, then 𝑃 𝐸 = 1 − 𝑃(𝐸 ′ ).
%
=0.237
%
The complement of
=0.371
‘has party affiliation’ is
7.2%=0.072
‘has no party affiliation’.
3.5%=0.035
4.4%=0.044
%
=0.241
MATH 110 Sec 13-2 Lecture: Operations With Events
For example, given the chart below about a group of voters’ party
affiliation, if the event 𝐸 is ‘Person has a party affiliation’, find P(𝐸).
The answer is
𝑃 ℎ𝑎𝑠 𝑎𝑓𝑓𝑖𝑙𝑖𝑎𝑡𝑖𝑜𝑛 = 𝑃 𝐷𝑒𝑚 + 𝑃 𝑅𝑒𝑝 + 𝑃 𝐺𝑟 + 𝑃 𝐿𝑖𝑏 + 𝑃 𝑂𝑡ℎ𝑒𝑟
𝑃 ℎ𝑎𝑠 𝑎𝑓𝑓𝑖𝑙𝑖𝑎𝑡𝑖𝑜𝑛 = 0.371 + 0.241 + 0.044 + 0.035 + 0.072 = 0.763
But it’s easier to use the previous result:
If 𝐸 is an event, then 𝑃 𝐸 = 1 − 𝑃(𝐸 ′ ).
%
=0.237
%
The complement of
=0.371
‘has party affiliation’ is
7.2%=0.072
‘has no party affiliation’.
3.5%=0.035
𝑃 ℎ𝑎𝑠 𝑎𝑓𝑓𝑖𝑙 = 1 − 𝑃 𝑛𝑜𝑛𝑒
4.4%=0.044
%
=0.241
MATH 110 Sec 13-2 Lecture: Operations With Events
For example, given the chart below about a group of voters’ party
affiliation, if the event 𝐸 is ‘Person has a party affiliation’, find P(𝐸).
The answer is
𝑃 ℎ𝑎𝑠 𝑎𝑓𝑓𝑖𝑙𝑖𝑎𝑡𝑖𝑜𝑛 = 𝑃 𝐷𝑒𝑚 + 𝑃 𝑅𝑒𝑝 + 𝑃 𝐺𝑟 + 𝑃 𝐿𝑖𝑏 + 𝑃 𝑂𝑡ℎ𝑒𝑟
𝑃 ℎ𝑎𝑠 𝑎𝑓𝑓𝑖𝑙𝑖𝑎𝑡𝑖𝑜𝑛 = 0.371 + 0.241 + 0.044 + 0.035 + 0.072 = 0.763
But it’s easier to use the previous result:
If 𝐸 is an event, then 𝑃 𝐸 = 1 − 𝑃(𝐸 ′ ).
%
=0.237
%
The complement of
=0.371
‘has party affiliation’ is
7.2%=0.072
‘has no party affiliation’.
3.5%=0.035
𝑃 ℎ𝑎𝑠 𝑎𝑓𝑓𝑖𝑙 = 1 − 𝑃 𝑛𝑜𝑛𝑒
𝑃 ℎ𝑎𝑠 𝑎𝑓𝑓𝑖𝑙 = 1 − 0.237
4.4%=0.044
%
=0.241
MATH 110 Sec 13-2 Lecture: Operations With Events
For example, given the chart below about a group of voters’ party
affiliation, if the event 𝐸 is ‘Person has a party affiliation’, find P(𝐸).
The answer is
𝑃 ℎ𝑎𝑠 𝑎𝑓𝑓𝑖𝑙𝑖𝑎𝑡𝑖𝑜𝑛 = 𝑃 𝐷𝑒𝑚 + 𝑃 𝑅𝑒𝑝 + 𝑃 𝐺𝑟 + 𝑃 𝐿𝑖𝑏 + 𝑃 𝑂𝑡ℎ𝑒𝑟
𝑃 ℎ𝑎𝑠 𝑎𝑓𝑓𝑖𝑙𝑖𝑎𝑡𝑖𝑜𝑛 = 0.371 + 0.241 + 0.044 + 0.035 + 0.072 = 0.763
But it’s easier to use the previous result:
If 𝐸 is an event, then 𝑃 𝐸 = 1 − 𝑃(𝐸 ′ ).
%
=0.237
%
The complement of
=0.371
‘has party affiliation’ is
7.2%=0.072
‘has no party affiliation’.
3.5%=0.035
𝑃 ℎ𝑎𝑠 𝑎𝑓𝑓𝑖𝑙 = 1 − 𝑃 𝑛𝑜𝑛𝑒 4.4%=0.044
𝑃 ℎ𝑎𝑠 𝑎𝑓𝑓𝑖𝑙 = 1 − 0.237 = 0.763
%
=0.241
MATH 110 Sec 13-2 Lecture: Operations With Events
For example, given the chart below about a group of voters’ party
affiliation, if the event 𝐸 is ‘Person has a party affiliation’, find P(𝐸).
The answer is
𝑃 ℎ𝑎𝑠 𝑎𝑓𝑓𝑖𝑙𝑖𝑎𝑡𝑖𝑜𝑛 = 𝑃 𝐷𝑒𝑚 + 𝑃 𝑅𝑒𝑝 + 𝑃 𝐺𝑟 + 𝑃 𝐿𝑖𝑏 + 𝑃 𝑂𝑡ℎ𝑒𝑟
𝑃 ℎ𝑎𝑠 𝑎𝑓𝑓𝑖𝑙𝑖𝑎𝑡𝑖𝑜𝑛 = 0.371 + 0.241 + 0.044 + 0.035 + 0.072 = 0.763
But it’s easier to use the previous result:
If 𝐸 is an event, then 𝑃 𝐸 = 1 − 𝑃(𝐸 ′ ).
%
=0.237
%
The complement of
=0.371
‘has party affiliation’ is
7.2%=0.072
‘has no party affiliation’.
3.5%=0.035
𝑃 ℎ𝑎𝑠 𝑎𝑓𝑓𝑖𝑙 = 1 − 𝑃 𝑛𝑜𝑛𝑒 4.4%=0.044
𝑃 ℎ𝑎𝑠 𝑎𝑓𝑓𝑖𝑙 = 1 − 0.237 = 0.763
%
=0.241
MATH 110 Sec 13-2 Lecture: Operations With Events
UNION RULE FOR PROBABILITIES
𝑃 𝐸 ∪ 𝐹 = 𝑃 𝐸 + 𝑃 𝐹 − 𝑃(𝐸 ∩ 𝐹)
MATH 110 Sec 13-2 Lecture: Operations With Events
UNION RULE FOR PROBABILITIES
𝑃 𝐸 ∪ 𝐹 = 𝑃 𝐸 + 𝑃 𝐹 − 𝑃(𝐸 ∩ 𝐹)
It is easy to see why this is true if we construct a Venn Diagram.
MATH 110 Sec 13-2 Lecture: Operations With Events
UNION RULE FOR PROBABILITIES
𝑃 𝐸 ∪ 𝐹 = 𝑃 𝐸 + 𝑃 𝐹 − 𝑃(𝐸 ∩ 𝐹)
It is easy to see why this is true if we construct a Venn Diagram.
MATH 110 Sec 13-2 Lecture: Operations With Events
UNION RULE FOR PROBABILITIES
𝑃 𝐸 ∪ 𝐹 = 𝑃 𝐸 + 𝑃 𝐹 − 𝑃(𝐸 ∩ 𝐹)
It is easy to see why this is true if we construct a Venn Diagram.
E
F
The subtraction is
needed because, as
you can see, 𝐸 ∩ 𝐹
gets included twice,
once because it’s
part of E and again
because it’s part of F.
MATH 110 Sec 13-2 Lecture: Operations With Events
UNION RULE FOR PROBABILITIES
𝑃 𝐸 ∪ 𝐹 = 𝑃 𝐸 + 𝑃 𝐹 − 𝑃(𝐸 ∩ 𝐹)
MATH 110 Sec 13-2 Lecture: Operations With Events
If we select a single card from a standard 52-card deck, what
is the probability that we draw either a heart or a face card?
MATH 110 Sec 13-2 Lecture: Operations With Events
If we select a single card from a standard 52-card deck, what
is the probability that we draw either a heart or a face card?
Let H be the event ‘draw a heart’ and F be ‘draw a face card’.
MATH 110 Sec 13-2 Lecture: Operations With Events
If we select a single card from a standard 52-card deck, what
is the probability that we draw either a heart or a face card?
Let H be the event ‘draw a heart’ and F be ‘draw a face card’.
That means that we are looking for 𝑃(𝐻 ∪ 𝐹).
MATH 110 Sec 13-2 Lecture: Operations With Events
If we select a single card from a standard 52-card deck, what
is the probability that we draw either a heart or a face card?
Let H be the event ‘draw a heart’ and F be ‘draw a face card’.
That means that we are looking for 𝑃(𝐻 ∪ 𝐹).
Recall what a standard 52-card deck looks like.
MATH 110 SecStandard
13-2 Lecture:
Operations
With Events
deck of
cards
4 suits (CLUBS, SPADES, HEARTS, DIAMONDS)
12 FaceCards
If we select a single card from a standard 52-card deck, what
is the probability that we draw either a heart or a face card?
Let H be the event ‘draw a heart’ and F be ‘draw a face card’.
That means that we are looking for 𝑃(𝐻 ∪ 𝐹).
Recall what a
13
HEARTS
standard
52-card deck looks like.
MATH 110 SecStandard
13-2 Lecture:
Operations
With Events
deck of
cards
4 suits (CLUBS, SPADES, HEARTS, DIAMONDS)
12 FaceCards
If we select a single card from a standard 52-card deck, what
UNION RULE FOR PROBABILITIES
is the probability that we draw either a heart or a face card?
𝑃 𝐻 ∪ 𝐹 = 𝑃 𝐻 + 𝑃 𝐹 − 𝑃(𝐻 ∩ 𝐹)
Let H be the
heart’
and
be ‘draw
13event
12 ‘draw
3 a13
+ 12 −
3 F 22
11 a face card’.
𝑃That
𝐻 ∪means
𝐹 = that
+ we
− are=looking for 𝑃(𝐻
= ∪=𝐹).
52 52 52
52
52 26
Recall what a
13
HEARTS
standard
52-card deck looks like.
MATH 110 SecStandard
13-2 Lecture:
Operations
With Events
deck of
cards
4 suits (CLUBS, SPADES, HEARTS, DIAMONDS)
12 FaceCards
If we select a single card from a standard 52-card deck, what
UNION RULE FOR PROBABILITIES
is the probability that we draw either a heart or a face card?
𝑃 𝐻 ∪ 𝐹 = 𝑃 𝐻 + 𝑃 𝐹 − 𝑃(𝐻 ∩ 𝐹)
Let H be the
heart’
and
be ‘draw
13event
12 ‘draw
3 a13
+ 12 −
3 F 22
11 a face card’.
𝑃That
𝐻 ∪means
𝐹 = that
+ we
− are=looking for 𝑃(𝐻
= ∪=𝐹).
52 52 52
52
52 26
Recall what a
13
HEARTS
standard
52-card deck looks like.
MATH 110 SecStandard
13-2 Lecture:
Operations
With Events
deck of
cards
4 suits (CLUBS, SPADES, HEARTS, DIAMONDS)
12 FaceCards
If we select a single card from a standard 52-card deck, what
UNION RULE FOR PROBABILITIES
is the probability that we draw either a heart or a face card?
𝑃 𝐻 ∪ 𝐹 = 𝑃 𝐻 + 𝑃 𝐹 − 𝑃(𝐻 ∩ 𝐹)
Let H be the
heart’
and
be ‘draw
13event
12 ‘draw
3 a13
+ 12 −
3 F 22
11 a face card’.
𝑃That
𝐻 ∪means
𝐹 = that
+ we
− are=looking for 𝑃(𝐻
= ∪=𝐹).
52 52 52
52
52 26
Recall what a
13
HEARTS
standard
52-card deck looks like.
MATH 110 SecStandard
13-2 Lecture:
Operations
With Events
deck of
cards
4 suits (CLUBS, SPADES, HEARTS, DIAMONDS)
12 FaceCards
If we select a single card from a standard 52-card deck, what
UNION RULE FOR PROBABILITIES
is the probability that we draw either a heart or a face card?
𝑃 𝐻 ∪ 𝐹 = 𝑃 𝐻 + 𝑃 𝐹 − 𝑃(𝐻 ∩ 𝐹)
Let H be the
heart’
and
be ‘draw
13event
12 ‘draw
3 a13
+ 12 −
3 F 22
11 a face card’.
𝑃That
𝐻 ∪means
𝐹 = that
+ we
− are=looking for 𝑃(𝐻
= ∪=𝐹).
52 52 52
52
52 26
Recall what a
13
HEARTS
standard
52-card deck looks like.
MATH 110 SecStandard
13-2 Lecture:
Operations
With Events
deck of
cards
4 suits (CLUBS, SPADES, HEARTS, DIAMONDS)
12 FaceCards
If we select a single card from a standard 52-card deck, what
UNION RULE FOR PROBABILITIES
is the probability that we draw either a heart or a face card?
𝑃 𝐻 ∪ 𝐹 = 𝑃 𝐻 + 𝑃 𝐹 − 𝑃(𝐻 ∩ 𝐹)
Let H be the
heart’
and
be ‘draw
13event
12 ‘draw
3 a13
+ 12 −
3 F 22
11 a face card’.
𝑃That
𝐻 ∪means
𝐹 = that
+ we
− are=looking for 𝑃(𝐻
= ∪=𝐹).
52 52 52
52
52 26
Recall what a
13
HEARTS
standard
52-card deck looks like.
MATH 110 SecStandard
13-2 Lecture:
Operations
With Events
deck of
cards
4 suits (CLUBS, SPADES, HEARTS, DIAMONDS)
12 FaceCards
If we select a single card from a standard 52-card deck, what
UNION RULE FOR PROBABILITIES
is the probability that we draw either a heart or a face card?
𝑃 𝐻 ∪ 𝐹 = 𝑃 𝐻 + 𝑃 𝐹 − 𝑃(𝐻 ∩ 𝐹)
Let H be the
heart’
and
be ‘draw
13event
12 ‘draw
3 a13
+ 12 −
3 F 22
11 a face card’.
𝑃That
𝐻 ∪means
𝐹 = that
+ we
− are=looking for 𝑃(𝐻
= ∪=𝐹).
52 52 52
52
52 26
Recall what a
13
HEARTS
standard
52-card deck looks like.
MATH 110 SecStandard
13-2 Lecture:
Operations
With Events
deck of
cards
4 suits (CLUBS, SPADES, HEARTS, DIAMONDS)
12 FaceCards
If we select a single card from a standard 52-card deck, what
UNION RULE FOR PROBABILITIES
is the probability that we draw either a heart or a face card?
𝑃 𝐻 ∪ 𝐹 = 𝑃 𝐻 + 𝑃 𝐹 − 𝑃(𝐻 ∩ 𝐹)
Let H be the
heart’
and
be ‘draw
13event
12 ‘draw
3 a13
+ 12 −
3 F 22
11 a face card’.
𝑃That
𝐻 ∪means
𝐹 = that
+ we
− are=looking for 𝑃(𝐻
= ∪=𝐹).
52 52 52
52
52 26
Recall what a
13
HEARTS
standard
52-card deck looks like.
Before
let’s note
a specialWith
case.
MATH
110 leaving
Sec 13-2this,
Lecture:
Operations
Events
UNION RULE FOR PROBABILITIES
𝑃 𝐸 ∪ 𝐹 = 𝑃 𝐸 + 𝑃 𝐹 − 𝑃(𝐸 ∩ 𝐹)
Before
let’s note
a specialWith
case.
MATH
110 leaving
Sec 13-2this,
Lecture:
Operations
Events
UNION RULE FOR PROBABILITIES
𝑃 𝐸 ∪ 𝐹 = 𝑃 𝐸 + 𝑃 𝐹 − 𝑃(𝐸 ∩ 𝐹)
UNION RULE FOR PROBABILITIES
WHEN E AND F ARE MUTUALLY EXCLUSIVE (𝐸 ∩ 𝐹 = ∅)
𝑃 𝐸 ∪ 𝐹 = 𝑃 𝐸 + 𝑃 𝐹 − 𝑃(𝐸 ∩ 𝐹)
E
F
If 𝐸 ∩ 𝐹 = ∅, then
P(𝐸 ∩ 𝐹) = 0
Before
let’s note
a specialWith
case.
MATH
110 leaving
Sec 13-2this,
Lecture:
Operations
Events
UNION RULE FOR PROBABILITIES
𝑃 𝐸 ∪ 𝐹 = 𝑃 𝐸 + 𝑃 𝐹 − 𝑃(𝐸 ∩ 𝐹)
UNION RULE FOR PROBABILITIES
WHEN E AND F ARE MUTUALLY EXCLUSIVE (𝐸 ∩ 𝐹 = ∅)
𝑃 𝐸∪𝐹 =𝑃 𝐸 +𝑃 𝐹 − 0
E
F
If 𝐸 ∩ 𝐹 = ∅, then
P(𝐸 ∩ 𝐹) = 0
MATH 110 Sec 13-2 Lecture: Operations With Events
UNION RULE FOR PROBABILITIES
𝑃 𝐸 ∪ 𝐹 = 𝑃 𝐸 + 𝑃 𝐹 − 𝑃(𝐸 ∩ 𝐹)
UNION RULE FOR PROBABILITIES
WHEN E AND F ARE MUTUALLY EXCLUSIVE (𝐸 ∩ 𝐹 = ∅)
𝑃 𝐸∪𝐹 =𝑃 𝐸 +𝑃 𝐹 − 0
E
F
MATH 110 Sec 13-2 Lecture: Operations With Events
UNION
RULEwhen
FOR PROBABILITIES
In other
words,
the two events are
mutually
to subtract.
𝑃 𝐸 ∪ exclusive,
𝐹 = 𝑃 𝐸there
+ 𝑃is nothing
𝐹 − 𝑃(𝐸
∩ 𝐹)
UNION RULE FOR PROBABILITIES
WHEN E AND F ARE MUTUALLY EXCLUSIVE (𝐸 ∩ 𝐹 = ∅)
𝑃 𝐸∪𝐹 =𝑃 𝐸 +𝑃 𝐹
E
F
MATH 110 Sec 13-2 Lecture: Operations With Events
UNION RULE FOR PROBABILITIES
𝑃 𝐸 ∪ 𝐹 = 𝑃 𝐸 + 𝑃 𝐹 − 𝑃(𝐸 ∩ 𝐹)
UNION RULE FOR PROBABILITIES
WHEN E AND F ARE MUTUALLY EXCLUSIVE (𝐸 ∩ 𝐹 = ∅)
𝑃 𝐸∪𝐹 =𝑃 𝐸 +𝑃 𝐹
MATH 110 Sec 13-2 Lecture: Operations With Events
A survey of consumers shows the amount of time they spend shopping
online per month compared to their annual income (see below). T is
the event ‘spends 10 or more hours/month shopping online’.
A is the event ‘has an annual income above $60,000’
Annual Income
10 + Hours (T)
3 – 9 Hours
0 – 2 Hours
Totals
Above $60,000 (A)
192
176
128
496
$40,000 - $60,000
160
208
144
512
Below $40,000
128
192
272
592
Totals
480
576
544
1,600
MATH 110 Sec 13-2 Lecture: Operations With Events
A survey of consumers shows the amount of time they spend shopping
online per month compared to their annual income (see below). T is
the event ‘spends 10 or more hours/month shopping online’.
A is the event ‘has an annual income above $60,000’
Annual Income
10 + Hours (T)
3 – 9 Hours
0 – 2 Hours
Totals
Above $60,000 (A)
192
176
128
496
$40,000 - $60,000
160
208
144
512
Below $40,000
128
192
272
592
Totals
480
576
544
1,600
What is the probability that a randomly-selected consumer neither shops
online 10 or more hours/month nor has an annual income above $60,000?
MATH 110 Sec 13-2 Lecture: Operations With Events
A survey of consumers shows the amount of time they spend shopping
online per month compared to their annual income (see below). T is
the event ‘spends 10 or more hours/month shopping online’.
A is the event ‘has an annual income above $60,000’
Annual Income
10 + Hours (T)
3 – 9 Hours
0 – 2 Hours
Totals
Above $60,000 (A)
192
176
128
496
$40,000 - $60,000
160
208
144
512
Below $40,000
128
192
272
592
Totals
480
576
544
1,600
What is the probability that a randomly-selected consumer neither shops
online 10 or more hours/month nor has an annual income above $60,000?
One thing that I hope that you learn with this material is that often you can
solve a probability problem without having to formally ‘use a formula’.
MATH 110 Sec 13-2 Lecture: Operations With Events
For example,
here the
problem
asks us of
to time
compute
𝑇 ∩shopping
𝐴 ′ ].
A survey
of consumers
shows
the amount
they𝑃[
spend
We could
apply the
complement
rule to (see
get below). T is
online per month
compared
to their
annual income
′ =hours/month
the event ‘spends
or𝐴more
𝑃 10
𝑇∪
1 − 𝑃(𝑇 ∪ 𝐴)shopping online’.
A isand
the then
eventthe
‘has
an annual
income
aboveto
$60,000’
Union
Rule for
Probability
get
𝑃 𝑇 Annual
∪ 𝐴 ′ Income
= 1 − 𝑃10 𝑇+ Hours
∪ 𝐴 (T)= 13 –−9 [𝑃
𝑇 +0 –𝑃2 Hours
𝐴 − 𝑃 Totals
𝑇∩𝐴 ]
Hours
butAbove
in reality,
think about
$60,000if(A)you just
192
176what is being
128 asked,
496you
can
solve
the problem
pulling the144appropriate
$40,000
- $60,000
160directly by
208
512
numbers
from the
table. 272
Below $40,000
128
192
592
Totals
480
576
544
1,600
What is the probability that a randomly-selected consumer neither shops
online 10 or more hours/month nor has an annual income above $60,000?
One thing that I hope that you learn with this material is that often you can
solve a probability problem without having to formally ‘use a formula’.
MATH 110 Sec 13-2 Lecture: Operations With Events
A survey of consumers shows the amount of time they spend shopping
online per month compared to theirThese
annual
(see below).
T is
areincome
the consumers
that don’t
the event ‘spends 10 or more hours/month
online’.
shop online 10shopping
or more hours/month.
A is the event ‘has an annual income above $60,000’
Annual Income
10 + Hours (T)
3 – 9 Hours
0 – 2 Hours
Totals
Above $60,000 (A)
192
176
128
496
$40,000 - $60,000
160
208
144
512
Below $40,000
128
192
272
592
Totals
480
576
544
1,600
What is the probability that a randomly-selected consumer neither shops
online 10 or more hours/month nor has an annual income above $60,000?
MATH 110 Sec 13-2 Lecture: Operations With Events
A survey of consumers shows the amount of time they spend shopping
online per month compared to theirThese
annual
(see below).
T is
areincome
the consumers
that don’t
These
are
the
consumers
that
don’t
the event ‘spends 10 or more hours/month
online’.
shop online 10shopping
or more hours/month.
have an annual
income
above
A is the
event
‘has$60,000.
an annual income above $60,000’
Annual Income
10 + Hours (T)
3 – 9 Hours
0 – 2 Hours
Totals
Above $60,000 (A)
192
176
128
496
$40,000 - $60,000
160
208
144
512
Below $40,000
128
192
272
592
Totals
480
576
544
1,600
What is the probability that a randomly-selected consumer neither shops
online 10 or more hours/month nor has an annual income above $60,000?
MATH 110 Sec 13-2 Lecture: Operations With Events
A survey of consumers shows the amount of time they spend shopping
online per month compared to theirThese
annual
(see below).
T is
areincome
the consumers
that don’t
These
are
the
consumers
that
don’t
the event ‘spends 10 or more hours/month
online’.
shop online 10shopping
or more hours/month.
have an annual
income
above
A is the
event
‘has$60,000.
an annual income above $60,000’
Annual Income
10 + Hours (T)
$60,000 (A)
192
So, theseAbove
are the
consumers that
$40,000
don’t
shop- $60,000
online 10 + 160
Below
$40,000
hours/month
nor
do they have128
an
Totals
480
annual income
above $60,000.
3 – 9 Hours
0 – 2 Hours
Totals
176
128
496
208
144
512
192
272
592
576
544
1,600
What is the probability that a randomly-selected consumer neither shops
online 10 or more hours/month nor has an annual income above $60,000?
MATH 110 Sec 13-2 Lecture: Operations With Events
A survey of consumers shows the amount of time they spend shopping
online per month compared to theirThese
annual
(see below).
T is
areincome
the consumers
that don’t
These
are
the
consumers
that
don’t
the event ‘spends 10 or more hours/month
online’.
shop online 10shopping
or more hours/month.
have an annual
income
above
A is the
event
‘has$60,000.
an annual income above $60,000’
Annual Income
10 + Hours (T)
$60,000 (A)
192
So, theseAbove
are the
consumers that
$40,000
don’t
shop- $60,000
online 10 + 160
Below
$40,000
hours/month
nor
do they have128
an
Totals
480
annual income
above $60,000.
3 – 9 Hours
0 – 2 Hours
Totals
176
128
496
208
144
512
192
272
592
576
544
1,600
What is the probability that a randomly-selected consumer neither shops
online 10 or more hours/month nor has an annual income above $60,000?
208 + 144 + 192 + 272
816
𝑑𝑜𝑛′ 𝑡 𝑠ℎ𝑜𝑝 10 + ℎ𝑟𝑠 𝑎 𝑚𝑜𝑛𝑡ℎ 𝑜𝑛𝑙𝑖𝑛𝑒
𝑃
=
=
= 0.51
𝑛𝑜𝑟 ℎ𝑎𝑣𝑒 𝑖𝑛𝑐𝑜𝑚𝑒 𝑎𝑏𝑜𝑣𝑒 $60,000
1600
1600
MATH 110 Sec 13-2 Lecture: Operations With Events
A survey of consumers shows the amount of time they spend shopping
online per month compared to theirThese
annual
(see below).
T is
areincome
the consumers
that don’t
These
are
the
consumers
that
don’t
the event ‘spends 10 or more hours/month
online’.
shop online 10shopping
or more hours/month.
have an annual
income
above
A is the
event
‘has$60,000.
an annual income above $60,000’
Annual Income
10 + Hours (T)
$60,000 (A)
192
So, theseAbove
are the
consumers that
$40,000
don’t
shop- $60,000
online 10 + 160
Below
$40,000
hours/month
nor
do they have128
an
Totals
480
annual income
above $60,000.
3 – 9 Hours
0 – 2 Hours
Totals
176
128
496
208
144
512
192
272
592
576
544
1,600
What is the probability that a randomly-selected consumer neither shops
online 10 or more hours/month nor has an annual income above $60,000?
208 + 144 + 192 + 272
816
𝑑𝑜𝑛′ 𝑡 𝑠ℎ𝑜𝑝 10 + ℎ𝑟𝑠 𝑎 𝑚𝑜𝑛𝑡ℎ 𝑜𝑛𝑙𝑖𝑛𝑒
𝑃
=
=
= 0.51
𝑛𝑜𝑟 ℎ𝑎𝑣𝑒 𝑖𝑛𝑐𝑜𝑚𝑒 𝑎𝑏𝑜𝑣𝑒 $60,000
1600
1600
MATH 110 Sec 13-2 Lecture: Operations With Events
A survey of consumers shows the amount of time they spend shopping
online per month compared to theirThese
annual
(see below).
T is
areincome
the consumers
that don’t
These
are
the
consumers
that
don’t
the event ‘spends 10 or more hours/month
online’.
shop online 10shopping
or more hours/month.
have an annual
income
above
A is the
event
‘has$60,000.
an annual income above $60,000’
Annual Income
10 + Hours (T)
$60,000 (A)
192
So, theseAbove
are the
consumers that
$40,000
don’t
shop- $60,000
online 10 + 160
Below
$40,000
hours/month
nor
do they have128
an
Totals
480
annual income
above $60,000.
3 – 9 Hours
0 – 2 Hours
Totals
176
128
496
208
144
512
192
272
592
576
544
1,600
What is the probability that a randomly-selected consumer neither shops
online 10 or more hours/month nor has an annual income above $60,000?
208 + 144 + 192 + 272
816
𝑑𝑜𝑛′ 𝑡 𝑠ℎ𝑜𝑝 10 + ℎ𝑟𝑠 𝑎 𝑚𝑜𝑛𝑡ℎ 𝑜𝑛𝑙𝑖𝑛𝑒
𝑃
=
=
= 0.51
𝑛𝑜𝑟 ℎ𝑎𝑣𝑒 𝑖𝑛𝑐𝑜𝑚𝑒 𝑎𝑏𝑜𝑣𝑒 $60,000
1600
1600
MATH 110 Sec 13-2 Lecture: Operations With Events
A survey of consumers shows the amount of time they spend shopping
online per month compared to theirThese
annual
(see below).
T is
areincome
the consumers
that don’t
These
are
the
consumers
that
don’t
the event ‘spends 10 or more hours/month
online’.
shop online 10shopping
or more hours/month.
have an annual
income
above
A is the
event
‘has$60,000.
an annual income above $60,000’
Annual Income
10 + Hours (T)
$60,000 (A)
192
So, theseAbove
are the
consumers that
$40,000
don’t
shop- $60,000
online 10 + 160
Below
$40,000
hours/month
nor
do they have128
an
Totals
480
annual income
above $60,000.
3 – 9 Hours
0 – 2 Hours
Totals
176
128
496
208
144
512
192
272
592
576
544
1,600
What is the probability that a randomly-selected consumer neither shops
online 10 or more hours/month nor has an annual income above $60,000?
208 + 144 + 192 + 272
816
𝑑𝑜𝑛′ 𝑡 𝑠ℎ𝑜𝑝 10 + ℎ𝑟𝑠 𝑎 𝑚𝑜𝑛𝑡ℎ 𝑜𝑛𝑙𝑖𝑛𝑒
𝑃
=
=
= 0.51
𝑛𝑜𝑟 ℎ𝑎𝑣𝑒 𝑖𝑛𝑐𝑜𝑚𝑒 𝑎𝑏𝑜𝑣𝑒 $60,000
1600
1600
MATH 110 Sec 13-2 Lecture: Operations With Events
A survey of consumers shows the amount of time they spend shopping
online per month compared to theirThese
annual
(see below).
T is
areincome
the consumers
that don’t
These
are
the
consumers
that
don’t
the event ‘spends 10 or more hours/month
online’.
shop online 10shopping
or more hours/month.
have an annual
income
above
A is the
event
‘has$60,000.
an annual income above $60,000’
Annual Income
10 + Hours (T)
$60,000 (A)
192
So, theseAbove
are the
consumers that
$40,000
don’t
shop- $60,000
online 10 + 160
Belowor
$40,000
128
hours/month
have an annual
Totals $60,000. 480
income above
3 – 9 Hours
0 – 2 Hours
Totals
176
128
496
208
144
512
192
272
592
576
544
1,600
What is the probability that a randomly-selected consumer neither shops
online 10 or more hours/month nor has an annual income above $60,000?
208 + 144 + 192 + 272
816
𝑑𝑜𝑛′ 𝑡 𝑠ℎ𝑜𝑝 10 + ℎ𝑟𝑠 𝑎 𝑚𝑜𝑛𝑡ℎ 𝑜𝑛𝑙𝑖𝑛𝑒
𝑃
=
=
= 0.51
𝑛𝑜𝑟 ℎ𝑎𝑣𝑒 𝑖𝑛𝑐𝑜𝑚𝑒 𝑎𝑏𝑜𝑣𝑒 $60,000
1600
1600
MATH 110 Sec 13-2 Lecture: Operations With Events
A survey of consumers shows the amount of time they spend shopping
online per month compared to theirThese
annual
(see below).
T is
areincome
the consumers
that don’t
These
are
the
consumers
that
don’t
the event ‘spends 10 or more hours/month
online’.
shop online 10shopping
or more hours/month.
have an annual
income
above
A is the
event
‘has$60,000.
an annual income above $60,000’
Annual Income
10 + Hours (T)
$60,000 (A)
192
So, theseAbove
are the
consumers that
$40,000
don’t
shop- $60,000
online 10 + 160
Belowor
$40,000
128
hours/month
have an annual
Totals $60,000. 480
income above
3 – 9 Hours
0 – 2 Hours
Totals
176
128
496
208
144
512
192
272
592
576
544
1,600
What is the probability that a randomly-selected consumer neither shops
online 10 or more hours/month nor has an annual income above $60,000?
208 + 144 + 192 + 272
816
𝑑𝑜𝑛′ 𝑡 𝑠ℎ𝑜𝑝 10 + ℎ𝑟𝑠 𝑎 𝑚𝑜𝑛𝑡ℎ 𝑜𝑛𝑙𝑖𝑛𝑒
𝑃
=
=
= 0.51
𝑛𝑜𝑟 ℎ𝑎𝑣𝑒 𝑖𝑛𝑐𝑜𝑚𝑒 𝑎𝑏𝑜𝑣𝑒 $60,000
1600
1600
MATH 110 Sec 13-2 Lecture: Operations With Events
A survey of consumers shows the amount of time they spend shopping
online per month compared to theirThese
annual
(see below).
T is
areincome
the consumers
that don’t
These
are
the
consumers
that
don’t
the event ‘spends 10 or more hours/month
online’.
shop online 10shopping
or more hours/month.
have an annual
income
above
A is the
event
‘has$60,000.
an annual income above $60,000’
Annual Income
10 + Hours (T)
$60,000 (A)
192
So, theseAbove
are the
consumers that
$40,000
don’t
shop- $60,000
online 10 + 160
Belowor
$40,000
128
hours/month
have an annual
Totals $60,000. 480
income above
3 – 9 Hours
0 – 2 Hours
Totals
176
128
496
208
144
512
192
272
592
576
544
1,600
What is the probability that a randomly-selected consumer neither shops
online 10 or more hours/month nor has an annual income above $60,000?
208 + 144 + 192 + 272
816
𝑑𝑜𝑛′ 𝑡 𝑠ℎ𝑜𝑝 10 + ℎ𝑟𝑠 𝑎 𝑚𝑜𝑛𝑡ℎ 𝑜𝑛𝑙𝑖𝑛𝑒
𝑃
=
=
= 0.51
𝑛𝑜𝑟 ℎ𝑎𝑣𝑒 𝑖𝑛𝑐𝑜𝑚𝑒 𝑎𝑏𝑜𝑣𝑒 $60,000
1600
1600
MATH 110 Sec 13-2 Lecture: Operations With Events
A survey of consumers shows the amount of time they spend shopping
online per month compared to theirThese
annual
(see below).
T is
areincome
the consumers
that don’t
These
are
the
consumers
that
don’t
the event ‘spends 10 or more hours/month
online’.
shop online 10shopping
or more hours/month.
have an annual
income
above
A is the
event
‘has$60,000.
an annual income above $60,000’
Annual Income
10 + Hours (T)
$60,000 (A)
192
So, theseAbove
are the
consumers that
$40,000
don’t
shop- $60,000
online 10 + 160
Belowor
$40,000
128
hours/month
have an annual
Totals $60,000. 480
income above
3 – 9 Hours
0 – 2 Hours
Totals
176
128
496
208
144
512
192
272
592
576
544
1,600
What is the probability that a randomly-selected consumer neither shops
online 10 or more hours/month nor has an annual income above $60,000?
208 + 144 + 192 + 272
816
𝑑𝑜𝑛′ 𝑡 𝑠ℎ𝑜𝑝 10 + ℎ𝑟𝑠 𝑎 𝑚𝑜𝑛𝑡ℎ 𝑜𝑛𝑙𝑖𝑛𝑒
𝑃
=
=
= 0.51
𝑛𝑜𝑟 ℎ𝑎𝑣𝑒 𝑖𝑛𝑐𝑜𝑚𝑒 𝑎𝑏𝑜𝑣𝑒 $60,000
1600
1600