Probability & Venn diagrams

Probability & Venn diagrams

RecapEvent - 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 ExclusiveIf A and B are Mutually exclusive then either :

A can happen orB 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 diagramsThese 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

AA’

1 3 5 2 4 6

A B

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

A B

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

A B

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

A B

A B

A and B are Events

A and B are not Mutually exclusive as they overlap

Probabiliity (A or B) is A B

P(A u B) = Green space - ÷ Total Space

A B P(A u B) = P(A) + P(B) - P(A n B)

ExampleThe 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) Getting a pay riseb) Not getting a pay risec) Getting both a pay rise and pension benefitsd) Getting no pay rise or benefit increasee) Getting a pay rise or benefits

Step 1 – Fill in the Venn diagram

Let A~ Pay rise B~ Benefits

173A B

138 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 ProbabilityThese 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

A BBUT, 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

A B

And so Our Probability P(A|B) is the ratio of Green Space ÷ Red space

A B)(

)()|(

BP

BAPBAP

Example 1P(A)=0.3 ; P(B)= 0.4 and P(A|B)=0.5

Find1- 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

A B

=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 BP(A n B’)

0.1 0.20.5

0.2

+

Example 2One 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 Electrical62 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

Let M-mechanical E-electrical

c) Let F~ Fail [= 1- P(M n E)]

so we want

P(M’ | F) = P(E|F) = P(E n F) / P(F)

We want to find

a) P(M and E) ~ P(M n E) - passes both mechanical and electrical

In language of Probability we need to find

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

Given P(E) = 50/100 = 0.5P(M) = 62/100 = 0.62P(E n M’) = 15/100 = 0.15

E M0.15

0.350.27

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)=0.15P(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