Transcript Probability - Venn diagrams and Conditional Probabilities

Probability & Venn diagrams
Recap
Event - An event is the situation in which
we are interested
Probability- Is the chance of that event
happening
Outcome – Is what happens (result of experiment)
Prob (Event ) =
Number of ways of event happening
Total possible outcomes
Prob (Not Event) =
1- Prob(Event)
Mutually Exclusive
If A and B are Mutually exclusive then either :
A can happen or
B can happen,
but both can not happen at the same time
So P(A) + P(B) =1
If A and B are not Mutually exclusive then they can
both happen at the same time
So P(A) + P(B) ≠ 1
Venn diagrams
These are an excellent way of representing a
Probability space
We can use them to clearly represent a
situation and to calculate corresponding
probabilities
A dice is rolled and an
even number is
obtained, show this in
a Venn diagram
A~ event even number
A’
13 5
A
246
A and B are Events
A
B
A and B are not Mutually exclusive as they overlap
Probabliity (A) is
A
B
P(A) = Blue space ÷ Total Space
A and B are Events
A
B
A and B are not Mutually exclusive as they overlap
Probabliity (B) is
A
B
P(B) = Green space ÷ total Space
A and B are Events
A
B
A and B are not Mutually exclusive as they overlap
Probabliity (A and B) is
A
B
P(A n B ) = Black space ÷ Total Space
A and B are Events
A
B
A and B are not Mutually exclusive as they overlap
Probabiliity (A or B) is

P(A u B) =
Space
A
Green space -
B
A
B
P(A u B) = P(A) + P(B) - P(A n B)

÷ Total
Example
The manager of a factory claims that among his 400
employees:
‾ 312 got a pay rise last year
‾ 248 got increased pension benefits last year
‾ 173 got both pension benefits and pay rise last year
‾ 13 got neither
Using last years figures as your guide to this years prospects,
calculate the probability of:
a)
b)
c)
d)
e)
Getting a pay rise
Not getting a pay rise
Getting both a pay rise and pension benefits
Getting no pay rise or benefit increase
Getting a pay rise or benefits
Step 1 – Fill in the Venn diagram
A
138
173
Let A~ Pay rise
B~ Benefits
B
75
13
P(A) = (138+173) ÷ (138+173+75+13)
= 311/400 = 0.7775
P(not A) = P(A’) = 1- 311/400 = 0.2225
P(A n B) = 173/400 = 0.4325 - pay rise and benefits
P(A’ U B’) = 13/400 = 0.0325 - no rise or benefits
Conditional Probability
These are the probabilities calculated on the basis that
something has already happened
For example :
—The probability that I will pay my electricity bill given
that have just been paid
—The probability that my students will turn upto class
given that it is a sunny day
The emphasis is that the probability is influenced by
something that has already happened.
If these two events are A and B then they are not
INDEPENDENT we write P(A|B) ~ P(A given B)
P(A|B) ~ P(A given B has occurred)
If B has already happened then our event must be somewhere in B
B
A
BUT, How can A happen if our event must be in the B space ?
We can only be in the following Space on our Venn Diagram
B
A
And so Our Probability P(A|B) is the ratio of Green Space ÷ Red space
P( A  B)
P( A | B) 
P ( B)
A
B
Example 1
P(A)=0.3 ; P(B)= 0.4 and P(A|B)=0.5
Find
1- P(A n B)
2- P(A u B)
3- P(A|B’ ) Fill in the Venn diagram
1- P(A|B) is 0.5 so
=0.5
+
A
0.1
0.2
B
0.2
0.5
P(A|B) = P(A n B)/P(B) so
P(A n B) = 0.5 x 0.4 = 0.2
2- P( A u B) = P(A) + P(B) – P(A n B) = 0.3 + 0.4 – 0.2 = 0.5
3- P(A/B’) ~ P( A given Not B) = P(A n B’)/P(B’)
= (0.3-0.2)/(1-0.4)
= 0.1/0.6 = 0.1667
A
P(A n B’)
B
Example 2
One hundred cars are entered for an MOT test. The test comprises
two parts : Mechanical and electrical
The car must pass both parts to be given an MOT certificate.
Half the cars pass the Electrical
62 pass the Mechanical test
15 pass the Electrical but fail the electrical
Find the probability that a car chosen at random
a) Passes overall (i.e passes both tests)
b) Fails on one test only
c) Given that it had failed, failed the Mechanical
only
Draw a Venn diagram
We want to find
Let M-mechanical E-electrical
In language of Probability we need to find
a) P(M and E) ~ P(M n E) - passes both mechanical and electrical
b) P(M or E)
~ P( M u B) - passes mechanical or electrical or both
So we want
P(M’ or E’) ~ P( M’ u B’) - fails mechanical or electrical or both
c) Let F~ Fail [= 1- P(M n E)]
so we want
P(M’ | F) = P(E|F) = P(E n F) / P(F)
Given
P(E) = 50/100
P(M) = 62/100
P(E n M’) = 15/100
= 0.5
= 0.62
= 0.15
E
0.15
0.35
M
0.27
a) 100 cars and 35 pass overall
so
P(E n M) = 0.35
b) P(E u M) = P(E) + P(M) – P(E n M) = 0.5 +0.65 – 0.35 = 0.75
c) Given the car fails what chance that it failed Mechanical
only
P(En F)=0.15
P(F) = 1-P(Pass) = 1-0.35 = 0.65
P(M’|F) = P(E|F) = P(E n F)/P(F) = 0.15/0.65
= 15/65 = 0.2307