1. Probability The Language of Sets
Statistics 1
Probability
The Language of Sets

2. A little about John Venn and Venn diagrams before we start . . . .
John Venn was born in 1834 at Hull, Yorkshire. His mother, Martha Sykes, came from Swanland, near Hull.
Venn's main area of interest was logic and he published three texts on the subject. The second was
Symbolic Logic in which he introduced the Venn diagrams in 1881.
(4 August 1834 – 4 April 1923),

3. Venn diagrams
Consider drawing a diagram to show 20 people of whom there are 8 beer drinkers, 5 wine drinkers and 3 people who drink both. We can show this with the circles!
Beer
Wine 10
Beer
Wine
Easy eh!
Represents the 10 people who don’t drink beer or wine 5 2 3

4. 10
Beer
Wine 5 2 3
And now lets add the maths language, easy stuff first
If we choose a person at random the probability they drink beer is
P(beer) = P(wine)=
The area where beer and wine overlap is written as B ∩ W, and read as B intersection W. The probability that someone drinks wine and beer is written as P(B∩W), where
P(B∩W)=

5. 10
Beer
Wine 5 2 3
The probability that someone drinks wine or beer is written as P(BUW), and is read as B union W, where
P(BUW)=

6. 10
Beer
Wine 5 2 3
We can also find the probability that somebody doesn’t drink beer.
The notation is P(B') and is known as the complementary probability
P(B')=
Don’t forget P(B') = 1 – P(B)

7. 10
Beer
Wine 5 2 3
How do we write the probability of a person not drinking beer but drinking wine?
P(B'∩W) =

8. P(AUB) = P(A) + P(B) – P(A∩B)
Addition Rule for Probability A B
The diagram shows two events A and B.
The shaded area represents P(AUB), i.e. the probability of A or B occurring.
P(AUB) = P(A) + P(B) – P(A∩B)
Subtract because the members of the intersection are already included in P(A) and P(B)
IMPORTANT RULE

9. Mutually Exclusive Events
spade
red
A card is chosen from a pack of 52 playing cards. If that card is a spade then it cannot be a red card.
where two events can not occur at the same time they are said to be mutually exclusive so we have:
P(AUB) = P(A) + P(B)
If P(AUB) = 1, then the events A and B are said to be exhaustive (all the possible outcomes are covered)

10. Product Rule for Independent Events
If C is the event ‘result of spinning a fair coin’ and D is the event ‘number obtained when a fair dice is rolled’, then these events are independent and . . .
P(C ∩ D) = P(C) x P(D)
You can show this with a sample space diagram (probability space diagram)
Let C be the event ‘ a head is obtained’ and let D be the event ‘a number greater than 4’ occurs. H X T 1 2 3 4 5 6

11. H X T 1 2 3 4 5 6 From the diagram we can see that
P (C) = ½ , P (D) = ⅓ and P( C ∩ D) =
Using the product rule gives
P (C ∩ D) = P(C) x P(D)
= ½ x ⅓ =

12. Conditional Probability
Consider a group of students studying maths and chemistry as illustrated in the Venn diagram. C M
15 students study chemistry of which 8 also study maths. 7 12 8 73
The probability that a student studies maths given that they study chemistry, is
number of students in M ∩ C =
number of students in C

13. This is known as conditional probability and is written P(M|C).
It is read as ‘the probability of M given C’
It is possible to rearrange this formula to give:
P(M∩C) = P(M|C) x P(C)

14. P( multiple of 4 ∩ greater than 15)
Example
A card is chosen at random from a set of twenty-five cards numbered 1 to 25.
What is the probability that the card chosen is a multiple of 4, given that it is greater than 15?
P( multiple of 4 ∩ greater than 15)
P ( greater than 15)
P( multiple of 4|greater than 15) =
Three numbers are multiples of 4 and greater than 15 i.e. 16, 20, 24.
So P( multiple of 4 ∩ greater than 15) = P ( greater than 15) =
Giving P( multiple of 4|greater than 15) =