Probability of Mutually Exclusive
Events.
• A ~ You are at school
• B ~ You are bowling
• A∩B ~ You are at school and bowling
S
• A∩B = Ø (you can’t bowl at school)
2 Sets are disjoint if their
intersection is the empty set Ø
• Disjoint Sets
B
e.g. There is no bowling alley at school so you can not
both be bowling and at school.
Probability:
• P(A U B)=P(A)+P(B)
• In the school and bowling example we can
simply add the probabilities.
• We are not concerned with overlap
because these events are:
“Mutually Exclusive”.
• The equation above only works when
events are mutually exclusive.
EXAMPE
• You are at school 7x5=35 hours out of 168
hours in a week.
P( at School)=35/168=21%
• You bowl for 2 hours once a month
p( bowling) = 2/(24x30)= 0.003 or 0.3%
• P( School U bowling)= 21% + 0.3% = 21.3%
Text Book
• Your text book refers to the set A U B
as A or B.
This is the union of two sets and is meant to
be interpreted as one or the other or both.
When sets A and B are mutually exclusive
we can remove the ‘both’ condition from
the sentence above.
When Events are NOT Mutually
Exclusive
• Just as with counting when we subtract
the number of elements in the intersection:
n( A UB) = n( A) + n( B) – n( A∩B )
• We subtract the probability of the
intersection:
P( A UB) = P( A) + P( B) – P( A∩B )
EXAMPLE 2
• You spend 7 hours a week having lunch, 5
of those lunch hours are at school.
• What’s P( At School U Eating Lunch) ?
EXAMPLE 2
• You spend 7 hours a week having lunch, 5 of
those lunch hours are at school.
• What’s P( At School U Eating Lunch) ?
• A ~ At School
B ~ Eating Lunch
• P(A)=35/168=21% P(B)=7/168=4%
• P( A∩B ) = 5/168=3%
• P(AUB) = P(A)+P(B)-P(A∩B)
= 21% + 4% - 3% = 22%
We subtract P(A∩B) so that we don’t count the
intersection twice.
n( A UB) = n( A) + n( B) – n( A∩B )
• e.g. A = set of DM Students
B = set of P.E. Students
There are 28 DM students and 25 PE students.
12 are in both. How many students in total are
in DM or PE (or both)?
n(A)=28, n(B)=25, n( A∩B )=12
n( A UB) = n( A) + n( B) – n( A∩B )
=28 + 25 – 12
= 41
Example 3
Of the 120 students in a class, 30 speak
Chinese, 50 speak Spanish, 75 speak
French, 12 speak Spanish and Chinese,
30 speak Spanish and French, and 15
speak Chinese and French. Seven
students speak all three languages. What
is the probability that a randomly chosen
student speaks none of these languages?
7 speak all 3
15 speak Chinese and French
15
30 speak Spanish and French
12 speak Spanish and Chinese
23
5
50 speak Spanish
7
10
37
8
30 speak Chinese
75 Speak French
120 – 105 = 15
don’t speak any.
• P( a random student doesn’t speak any)
• = 15/120
• =12.5 %
• We used a Venn Diagram because it is
better suited for tracking the intersections
of sets in this case.
Adding Intersections With an Event
and it’s compliment.
• Often when dealing with more than one event it is useful
to use:
• P(A) = P(A∩B)+P(A∩B’)
• Example: Only parts that are not obviously defective get
shipped. Consider events:
• G~part is good, OD~part is obviously defective, OD’ ~
part is not obviously defective
• P(G) = P(G∩OD)+P(G∩OD’)
• P(G) = 0+P(G∩OD’)
• P(G) = P(G∩OD’) (this intersection can now be used to
calculate conditional probability.
Example 3.14 (an old test question)
• 11)You are trying to find a popular toy for your little sister for
Christmas. If you go to Toys R Them, there is a 66%
probability of finding it. If you go to FloorMart there is an
80% chance of you finding it.
• a) If you only have time to go to one store and you decide
which one by tossing a coin, what is the probability that you
find the toy for your sister? (4 marks)
setTheoryProbExample.notebook
• b) If you have a 60% chance of going to Toys R Them and
also have a 60% chance of going to FloorMart (regardless of
whether or not you went to Toys R Them) what is the
probability that you will find the toy for your sister? (4 marks)
setTheoryProbExample.notebook
Homework
• Page 340 Questions 1 to 9