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LBSRE1021 Data Interpretation. Lecture 4 Probability. Objectives. Explain the concept of probability Apply simple laws of probability Construct and use a tree diagram Construct and use a probability table. Probability Scale. How is a probability determined? (1).

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## LBSRE1021 Data Interpretation

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**LBSRE1021 Data Interpretation**Lecture 4 Probability**Objectives**• Explain the concept of probability • Apply simple laws of probability • Construct and use a tree diagram • Construct and use a probability table**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?**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**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**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**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**Laws of Probability**Special Law of Addition. Two events E1 and E2, The probability that either E1 occurs or E2 occurs is P(E1 or E2) = P(E1) + P(E2) Provided that E1 and E2 are mutually exclusive**Special Law of Addition**Example draw a card from a pack E1 = card is a heart, P(E1) = 1/4 E2 = card is a diamond P(E2) = 1/4 P(E1 or E2) = 1/4 + 1/4 = 1/2**Joint Probability**For two events E1 and E2, the probability they both occur is: P(E1 and E2) = P(E1) x P(E2) Provided the events are independent I.e. the outcome of E1 does not affect the outcome of E2**Joint Probability**Example Draw card from pack E1 = card is a heart P(E1) = 1/4 E2 = card is an ace P(E2) = 1/13 P(E1 and E2) = 1/4 x 1/13 = 1/52 I.e. card is ace of hearts. If events are independent they cannot be mutually exclusive**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**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**Tabular Data (3)**(c) bought a large car, given that he previously had a small or medium car. 36 + 11 = 0.132 180 + 176

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