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Lecture Goals You should be able to:
- independent events
Reading Chapter 8
Homework Chapter 8: 8-12,17-19, 23, 25
2 Approaches 1) a priori = deduced from reason 2) a posteriori = found through data collection and analyses a priori Probability of rolling a 6 on 1 die a posteriori # events possible = 1 . 6 Comes into play when you want the probability of this OR that Examples: roll 5 or 6 draw 5 or 6 from deck p(A or B) = p(A) + p(B) - p(A and B)
P(draw Ace or club)
a posteriori = p(A or B) = p(A) + p(B) - p(A and B) p(Ace or club)=p(Ace)+p(club)-p(Ace and club) p(Ace or club)= 4/52 + 13/52 - 1/52 = 16/52 p(Ace or club)= .308 This way don't count the ace of clubs twice
Mutually Exclusive Sometimes events are mutually exclusive = both can't occur at same time Example: In 1 flip can't be both heads and tails at same time In 1 roll can't get both 5 and 6 at same time
Addition Rule When events are mutually exclusive: a posteriori = p(A or B) = p(A) + p(B) - p(A and B) When events are mutually exclusive: a posteriori = p(A or B) = p(A) + p(B) p(roll a 5 or a 6 in 1 roll) = p(roll 5) + p(roll 6) p(roll a 5 or a 6 in 1 roll) = 1/6+1/6 = 2/6 = .33
When events are mutually exclusive: a priori = p(roll a
5 or a 6 in 1 roll) = 2/6 = .33
Multiplication Rule Comes into play when you want the probability of this AND that on dice roll 5 and a 6 roll a 5 and draw 6 from deck Sometimes the occurrence of the first event effects the probability of the second = p(A and B) = p(A)p(B|A)a posteriori = p(A and B) = p(A)p(B|A) p(draw 10 then draw 4) = (1/52)(1/51) = .0004 p(draw 10 then draw 4) = (1/52)(1/51) = .0004 2 2 2 2 3 3 3 3 4 4 4 4 5 5 5 5 6 6 6 6 7 7 7 7 8 8 8 8 9 9 9 9 10 10 10 10 J J J J Q Q Q Q K K K K A A A A 2 2 2 2 3 3 3 3 4 4 4 4 5 5 5 5 6 6 6 6 7 7 7 7 8 8 8 8 9 9 9 9 X 10 10 10 J J J J Q Q Q Q K K K K A A A A a posteriori = p(A and
B) = p(A)p(B|A)
p(drawing 2 consecutive clubs) = (13/52)(12/51) = .059
When occurrence of 1st has no impact on occurrence of 2nd Random Sampling WITH replacement (draw a 10...put it back...draw a 10) Draw a 10 and roll a 6
a posteriori = p(A and B) = p(A)p(B|A) p(draw a 10 and roll a 6) = (4/52)(1/6) = .013
Multiplication & Addition Rules Some situations require using both rules to find solution a priori Die1) 1 1 1 1 1 1 2 2 2 2 2 2 3 3 3 3 3 3 4 4 4 4 4 4 5 5 5 5 5 5 6 6 6 6 6 6 Die2) 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6
1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5
6
P(dice total 11) = 2/36 = 1/18 = .055 a posteriori = p(roll a 5 and a 6 OR roll a 6 and a 5) a posteriori = p(roll a 5 and a 6) + p(roll a 6 and a 5)
p(roll a 5 and a 6) = (1/6)(1/6) = 1/36 p(roll a 6 and a 5) = p(roll a 6)p(roll a 5) p(roll a 6 and a 5) = (1/6)(1/6) = 1/36 a posteriori = p(roll a 5 and a 6) + p(roll a 6 and a 5) = 1/36 + 1/36 = 2/36 = .055
Ex - select from deck, roll dice
For continuous variables: p(A) = designated area under curve You study the population of sophomore women. Assume
the scores are normally distributed with m
= 120 pounds and s
= 8 pounds. If we randomly select one score (person) what is the p(score
³ 134)?
Table A area c for 1.75 = .0401 p(Xi ³
134) = 4.01%
Basic Probability and Experiments When a priori and a posteriori methods don't agree on same answer Example - How would you tell if a die is loaded? By logic each # rolled same percentage of times...so we roll 60 times and if fair..... 1) 10 times 2) 10 times 3) 10 times 4) 10 times 5) 10 times 6) 10 times
a posteriori = by calculation and data collection So we roll 60 times and count..... 1) 30 times 2) 6 times 3) 6 times 4) 6 times 5) 6 times 6) 6 times (What's expected) (What's observed) 1 10 30 2 10 6 3 10 6 4 10 6 5 10 6 6
10
6
Pattern shows the effect of loading the die Can use this idea to test the effect of ______ |
