1. LBSRE1021 Data Interpretation Lecture 4 Probability

2. Objectives Explain the concept of probability Apply simple laws of probability Construct and use a tree diagram Construct and use a probability table

3. Probability Scale

4. How is a probability determined? (1) 1. Subjective- estimate by experience 2. Empirical- by measurement p = No. times event occurred Total number of trials Affected by sampling error. E.g. toss a coin a number of times Is probability of heads 0.5?

5. How is a probability determined? (2) 3. A Priori Work out in advance Requires knowledge Assumes all outcomes equally likely e.g. probability of head 0.5 ace from pack of 52 cards 4/52 p= No. ways an event can occur Total number of possible outcomes

6. Pack of Cards 52 cards in pack Divided into 4 ‘suits’ –Clubs, Diamonds, Hearts, Spades 13 cards in each suit –Ace,2,3,4,5,6,7,8,9,10, Jack, Queen, King

7. Compound Events (1) Events Can be: –Independent: occurrence of one does not affect the other –Mutually Exclusive: either can occur but not both e.g. one card cannot be both Q and A Q and Heart not Mutually Exclusive

8. Compound Events (2) Mutually Exhaustive: set of all possible outcomes known The sum of the probabilities of a set of outcomes which are mutually exhaustive and mutually exclusive =1

9. Laws of Probability Special Law of Addition. Two events E 1 and E 2, The probability that either E 1 occurs or E 2 occurs is P(E 1 or E 2 ) = P(E 1 ) + P(E 2 ) Provided that E 1 and E 2 are mutually exclusive

10. Special Law of Addition Example draw a card from a pack E 1 = card is a heart, P(E 1 ) = 1/4 E 2 = card is a diamond P(E 2 ) = 1/4 P(E 1 or E 2 ) = 1/4 + 1/4 = 1/2

11. Joint Probability For two events E 1 and E 2, the probability they both occur is: P(E 1 and E 2 ) = P(E 1 ) x P(E 2 ) Provided the events are independent I.e. the outcome of E 1 does not affect the outcome of E 2

12. Joint Probability Example Draw card from pack E 1 = card is a heart P(E 1 ) = 1/4 E 2 = card is an ace P(E 2 ) = 1/13 P(E 1 and E 2 ) = 1/4 x 1/13 = 1/52 I.e. card is ace of hearts. If events are independent they cannot be mutually exclusive

13. General Law of Addition For two events E 1 and E 2, the probability that either E 1 occurs or E 2 or both occurs is: P(E 1 or E 2 or both) = P(E 1 ) + P(E 2 ) - [P(E 1 ) x P(E 2 )] ie events need not be mutually exclusive.

14. General Law of Addition Example E 1 = card is a heart, P(E 1 ) = 1/4 E 2 = card is an ace, P(E 2 ) = 1/13 P(E 1 or E 2 or both) = 1/4 + 1/13 - (1/4 x 1/13) = 16/52 = 4/13 Check: 16 possibilities - 13 hearts + 3 other aces

15. Conditional Probability (1) Outcome of a second event is conditional on the outcome of a first. Example Draw two cards from a pack. E 1 first card is an ace. E 2 second card is an ace What is P(E 2 )?

16. Conditional Probability (2) If first card is an ace, P(E 2 ) = 3/51 If first card is not an ace, P(E 2 ) = 4/51 Notation P(E 2 | E 1 ) = probability that E 2 occurs given that E1 has occurred. Probability that both events occur is P(E 1 and E 2 ) = P(E 2 | E 1 )

17. Conditional Probability (3) For drawing two aces P(E 1 ) = 4/52 = 1/13 P(E 2 | E 1 )= 3/51 P(E 2 and E 1 ) = 1/13 x 3/51 = 3/663 = 1/221 Conditional probability enables us to modify probability predictions in the light of any new or additional information.

18. Tree Diagrams The probability that machine A and machine B are still functioning in 5 year's time is 0.25 and 0.4 respectively. Find the probability that in 5 year's time (a)both are working (b)neither works (c)at least one machine works (d)just one machine is working

19. Tree Diagram Example

20. Tabular Data (1)

21. Tabular Data (2) Driver randomly selected. Find the probability that s/he (a)changed to a smaller car 47 + 22 + 69= 0.276 500 (b)changed to a larger car 36 + 11 + 63= 0.22 500

22. Tabular Data (3) (c)bought a large car, given that he previously had a small or medium car. 36 + 11 = 0.132 180 + 176