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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 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