LBSRE1021 Data Interpretation

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

100% Or 1.0 Certain

Likely

50% Or 0.5 Evens

Unlikely

0% Or 0.0 Impossible

How is a probability determined? (1)

1. Subjective- estimate by experience

2. Empirical- by measurementp = 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 PrioriWork 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

Tree Diagram Example

Machine A Machine B

Working 0.4 0.1

Working 0.25

Not working 0.6 0.15

Working 0.4 0.3

Not working 0.75

Not working 0.6 0.45

Tabular Data (1)

Present car size

Large Medium Small Total

Previous Large 75 47 22 144

car Medium 36 75 69 180

size Small 11 63 102 176

total 122 185 193 500

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