1. Society of Actuaries in Ireland
Probability 11 December 2017

2. Welcome to Society of Actuaries Tutorial on Probability
Introduction
What is Probability?
Question Focused Approach
2 hours - 10mins
Any questions? – Just ask your tutor!!
Thank you for coming along today
Introduce tutors
What is probability and why is it important
Go at your own pace
Encourage work together
The tutorial is going to last for 2 hours
Break in the middle
Tutors will be allocated to each group of students

3. Useful Resource for Project Maths
Normally €49.99
Discount code: SOA2018
€29.99 for the year
Access to 11 June 2018
Online exercises for active learning
Exam Technique guidance and video lessons on past exam papers

4. Combinations and Permutations Venn Diagrams Bernoulli Trials
Probability
Rules of Probability
Combinations and Permutations
Venn Diagrams
Bernoulli Trials
Expected Value

5. P(E) = 0 means that E can’t happen
Probability
Probability of event E = P(E) = 𝑁𝑜. 𝑜𝑓 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 𝑠𝑎𝑡𝑖𝑠𝑓𝑦𝑖𝑛𝑔 𝐸 𝑇𝑜𝑡𝑎𝑙 𝑛𝑜. 𝑜𝑓 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠
0 ≤ P(E) ≤ 1 meaning that probabilities must be between 0 and 1. Probabilities cannot be negative or greater than 1.
P(E) = 0 means that E can’t happen
P(E) = 1 means E is certain to happen
If you throw a die, what is the probability of getting an even number? 3/6
What is the probability of getting a 10? 0
What is the probability of getting less than 7? 1

6. Success/Failure
The sum of the probabilities of all possible mutually exclusive events is 1.
In particular, for the two mutually exclusive events success (A) or failure (A’)
𝑃 𝐴 +𝑃(𝐴′)=1
𝑃 𝐴 =1 −𝑃(𝐴′)
Or put another way:
𝑃 𝑆𝑢𝑐𝑐𝑒𝑠𝑠 +𝑃 𝑓𝑎𝑖𝑙𝑢𝑟𝑒 =1
𝑃 𝑆𝑢𝑐𝑐𝑒𝑠𝑠 =1 −𝑃(𝑓𝑎𝑖𝑙𝑢𝑟𝑒)
If for every possible age we know the probability that a student chosen at random will have that age, then these probabilities will sum to 1.
The two mutually exclusive events, success and failure, turn up frequently in exam questions.

7. Permutations and Combinations
Permutations is an arrangement of number of objects in a certain order, where the order is important.
Combinations is a selection of objects from a given set, where order is not important.

8. Permutations
The number of permutations (arrangements) of n different objects is n!
𝑛!=𝑛 𝑛−1 𝑛−2 … (2)(1)
For example, how many different arrangements of the letters of THURSDAY can I make. There are 8 letters, so the number of different ways they can be arranged is:
e.g. 8!=(8)(7)(6)(5)
Find the button on your calculator n!
8! =40,320

9. Permutations and Combinations
The number of permutations of r out of n different objects
𝑛! 𝑛−𝑟 !
e.g out of 8: (8) = 8! 8−5 !

10. Permutations and Combinations
The number of combinations (selections, where order does not matter), of r out of n different objects is given by
𝑛! 𝑟! 𝑛 −𝑟 ! = 𝑛 𝑛−1 𝑛−2 …(𝑛−𝑟+1) 𝑟! = nCr (Tables P 20)
e.g. how many combinations of 6 numbers can be selected from 42 numbers? n=42 r=6
42! 6! 42−6 !
For every combination there are r! permutations, so just divide the number of permutations by r!.
Open Tables at Page 20.
Answer is 5,245,786

11. If there are 100 students in the room and
Venn Diagram Example
If there are 100 students in the room and
90 use the tables (A)
13 use themathstutor.ie (B)
9 use both (A ∩B)
We can use this diagram to calculate probabilities.
If a student is chosen at random, what is the probability that they
use the tables
use the mathstutor.ie
use neither
use either or both?
Note: We could show the probabilities (.81, etc.) in the diagram in stead of the numbers in the sets. A B 4 81 9 6

12. Venn Diagram 𝐴
𝐴 ∩𝐵 𝐵
(𝐴∪𝐵)′
If an event can occur in one way A or in another way B this is denoted by 𝐴∪𝐵
including the possibility that both A and B occur together, this is denoted by 𝐴∩𝐵,
and also the possibility that neither occurs is denoted by (𝐴∪𝐵)′
Of the students in this room,
some use the approved formulae and tables (A),
some use themathstutor.ie (B).
How do you depict those who use both? (𝐴 ∩𝐵)
How do you depict those who use one or both? (𝐴∪𝐵)
How do you depict those who use neither? (𝐴∪𝐵)′
To find the probability that a student chosen at random will satisfy a particular event, divide the number who satisfy it by the total number of students in the room.

13. Addition Law (“or”) 𝐴
𝐴 ∩𝐵 𝐵
(𝐴∪𝐵)′
If an event can occur in one way A or in another way B (including the possibility of both A and B together)
𝑃 𝐴 𝒐𝒓 𝐵 =𝑃(𝐴∪𝐵)=𝑃 𝐴 + 𝑃 𝐵 −𝑃 𝐴∩𝐵
The −𝑃 𝐴∩𝐵 term is needed to avoid double counting.
In our example, students who use both the tables and themathstutor.ie would otherwise be double counted.

14. Addition Law (“or”) 𝐴 𝐵
(𝐴∪𝐵)′
If A and B are mutually exclusive (can’t happen together) then 𝑃 𝐴∩𝐵 =0 and
𝑃 𝐴 𝒐𝒓 𝐵 =𝑃(𝐴∪𝐵)=𝑃 𝐴 + 𝑃 𝐵
If event A is that a student is aged 17 and event B is that a student is aged 18, these two events are mutually exclusive, i.e. can’t happen together.
In a Venn diagram the ovals for mutually exclusive events do not overlap.
Remember that the simple form of the addition law only applies to mutually exclusive events.

15. Conditional Probability
If a student is chosen at random and we see that he has the tables then the probability that he uses themathstutor.ie is 9/90 = 0.09/0.90 = 0.10 B A
The prior probability that a student chosen at random uses themathstutor.ie is 13/100 = 0.13
but once we see that the chosen student has the tables we should use this information to narrow down the calculation:
There are now only 90 possible students of whom 9 use themathstutor.ie.
The probability that an event occurs given that another event has already occurred is called “conditional probability”. 81 9 4 6

16. Conditional Probability𝐴
𝐴 ∩𝐵 𝐵
(𝐴∪𝐵)′
The conditional probability that B occurs given A occurs
P 𝐵  𝐴 = 𝑃(𝐴 𝐵) 𝑃(𝐴)

17. Conditional Probability𝐴
𝐴 ∩𝐵 𝐵
(𝐴∪𝐵)′
The probability that both A and B occur is
𝑃 𝐴 𝑎𝑛𝑑 𝐵 =𝑃 𝐴 ∩𝐵 =𝑃 𝐴 𝑥 𝑃 𝐵 𝐴
also =𝑃 𝐵 𝑥 𝑃 𝐴 𝐵

18. Multiplication Law (“and”)𝐴
𝐴 ∩𝐵 𝐵
(𝐴∪𝐵)′
The events are independent if and only if
𝑃 𝐴 𝐵 = 𝑃 𝐴 and
𝑃 𝐵 𝐴 = 𝑃(𝐵) and
𝑃 𝐴 𝒂𝒏𝒅 𝐵 =𝑃 𝐴∩𝐵 =𝑃 𝐴 𝑥 𝑃(𝐵)
This is the definition of independence.
Remember that the simple form of the multiplication law only applies to independent events.

19. Bernouilli Trials (or Binomial Distribution)
Suppose we perform 6 trials with constant probability
p = P(success)
q = P(failure)
= 1 – p
in each trial, the outcomes being independent of each other.

20. Bernouilli Trials (or Binomial Distribution)
What is the probability of getting 4 successes?
One way of getting 4 successes would be to have success in the first 4 trials followed by 2 failures, with probability p4q2, using the Multiplication Law for independent events.

21. Bernouilli Trials
The number of ways that we can get 4 successes from 6 trials is 6C4
so the probability is 6C4 p4q2 , using the Addition Law for mutually exclusive events.
More generally
P(r successes) = nCr prqn-r (Tables Page 33)
nCr is the number of combinations of r out of n different objects, given by
nCr = 𝑛! 𝑟! 𝑛 −𝑟 ! = 𝑛 𝑛−1 𝑛−2 …(𝑛−𝑟+1) 𝑟!
This distribution is called the Binomial Distribution.
Open Tables pp 33, 20

22. Expected Values
X is a random variable. The probabiltiy that X takes on a particular value is P(X=x). The expected value or mean of X can be found using the following formula:
E(X) = ∑x*P(X=x)

23. Next Tutorial is on Monday 16 January 2018 On Calculus
Wrap Up
Next Tutorial is on Monday 16 January 2018
On Calculus
Back in Chartered Accountants House
6-8pm
Related to the topic of probability and statistics is Financial Maths. There are questions up on the website on this topic if you’re interested.
16th January – Calculus
23rd January – Numbers & Algebra with Functions
30th January – Revision