Probability

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Probability Venn diagrams
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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 (Not Event) 1- Prob(Event)
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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

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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
A
1 3 5
2 4 6
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A and B are Events
A and B are not Mutually exclusive as they
overlap Probabliity (A) is
A
B
P(A) Blue space Total Space
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A and B are Events
A and B are not Mutually exclusive as they
overlap Probabliity (B) is
A
B
P(B) Green space total Space
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A and B are Events
A and B are not Mutually exclusive as they
overlap Probabliity (A and B) is
P(A n B ) Black space Total Space
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A and B are Events
A and B are not Mutually exclusive as they
overlap Probabiliity (A or B) is
P(A u B) Green space -
Total Space
P(A u B) P(A) P(B) - P(A n B)
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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
• 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
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Let A Pay rise B Benefits
A
B
173
138
75
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P(A) (138173) (1381737513) 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
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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(AB) P(A given B)

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P(AB) P(A given B has occurred)
If B has already happened then our event must be
somewhere in B
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
And so Our Probability P(AB) is the ratio of
Green Space Red space
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Example 1
P(A)0.3 P(B) 0.4 and P(AB)0.5 Find 1- P(A
n B) 2- P(A u B) 3- P(AB ) Fill in the Venn
diagram
0.2
0.2
0.1
0.5
1- P(AB) is 0.5 so
P(AB) P(A n B)/P(B) so
P(A n B) 0.5 x 0.4 0.2
0.5

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
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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
• Passes overall (i.e passes both tests)
• Fails on one test only
• Given that it had failed, failed the Mechanical
  only
• Draw a Venn diagram

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Let M-mechanical E-electrical
We want to find
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(EF) P(E n F) /
P(F)
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Given
P(E) 50/100 0.5 P(M) 62/100
0.62 P(E n M) 15/100 0.15
E
M
0.35
0.15
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(MF) P(EF) P(E n
F)/P(F) 0.15/0.65 15/65 0.2307