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Praxis Weekend Analytics

Praxis Weekend Analytics

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- 1. Probability and Inferential Statistics • Probability – probability of occurrences are assigned in the inferential process under conditions of uncertainty
- 2. Methods of assigning probability • Classical method of assigning probability (rules and laws) • Relative frequency of occurrence (cumulated historical data) • Subjective Probability (personal intuition or reasoning)
- 3. Classical probability: Assigning probability • Each outcome is equally likely • Number of outcomes leading to the event divided by the total number of outcomes possible which is denoted as P(E) = ne/N where N = total number of outcomes, and ne = number of outcomes in event E P(E) = probability of event E
- 4. Classical probability: Assigning probability …. continued Probability of event E can also be defined as Number of outcomes favourable for E to occur divided by Total number of outcomes in Sample space S Measured on a scale between 0 and 1 inclusive. If E is impossible P(E) = 0, if E is certain then P(E) = 1.
- 5. Probability and related terms: Experiment and Trial • The process of observing a phenomenon that has variation in its outcomes. • The term is used in a broad sense to include any operation of data collection or observation where the outcomes are subject to variation. • Examples: rolling a die, drawing a card from a shuffled deck, sampling a number of customers for an opinion survey • Trial is the occurrence and any repetition of an experiment.
- 6. Probability and related terms: Sample space and Event • Sample space: The collection of all possible distinct outcomes of an experiment. Denoted by S. • Elementary outcome/ Simple event/ Element of the sample space: Each outcome of an experiment. • Event: The collection of elementary outcomes possessing a designated feature. An event A occurs when any one of the elementary outcomes in A occurs.
- 7. Four Types of Probability Marginal Union Joint Conditional P( X ) P( X Y ) P( X Y ) P( X| Y ) The probability of both X and Y occurring The probability of X occurring given that Y has occurred The probability of an event X occurring X The probability of X or Y or both occurring X Y X Y Y
- 8. General Law of Addition For any two events and , ( or ) = ( ∪ ) = ( ) + ( ) − ( ∩ ) + - =
- 9. General Law of Addition … Example P( N .56 S) P( N ) .70 S N .70 S ) P( N ) P ( S ) P ( N .67 P( S ) .67 P( N S ) .56 P( N S ) .70 .67 .56 0.81
- 10. Complement rule
- 11. Law of Multiplication • As we know, the intersection of two events is called the joint probability • General law of multiplication is used to find the joint probability • General law of multiplication gives the probability that both events x and y will occur at the same time • P(x|y) is a conditional probability that can be stated as the probability of x given y
- 12. Law of Conditional Probability • If A and B are two events, the conditional probability of A occurring given that B is known or has occurred is expressed as P(A|B) • The conditional probability of A given B is the joint probability of A and B divided by the marginal probability of B. P( A B) P( A | B) P( B) P( B | A) P( A) P( B)
- 13. Conditional Probability …….
- 14. Conditional Probability : Examples • P(card is a heart | it is a red suit) = ½ • P(Tomorrow is Tuesday | it is Monday) = 1 • P(result of coin toss is heads | the coin is fair) =1/2
- 15. Example: A batch of 5 computers has 2 faulty computers. If the computers are chosen at random (without replacement), what is the probability that the first two inspected are both faulty? Answer: P(first computer faulty AND second computer faulty)
- 16. Drawing cards Drawing two random cards from a pack without replacement, what is the probability of getting two hearts? [13 of the 52 cards in a pack are hearts] 1. 2. 3. 4. 1/16 3/51 3/52 1/4 46% 31% 14% 1 2 3 10% 4
- 17. Drawing cards Drawing two random cards from a pack without replacement, what is the probability of getting two hearts? To start with 13/52 of the cards are hearts. After one is drawn, only 12/51 of the remaining cards are hearts. So the probability of two hearts is
- 18. Throw of a die Throwing a fair dice, let events be A = get an odd number B = get a 5 or 6 What is P(A or B)?
- 19. Alternative “Probability of not getting either A or B = probability of not getting A and not getting B” = Complements Rule = i.e. P(A or B) = 1 – P(“not A” and “not B”)
- 20. Throw of a dice Throwing a fair dice, let events be A = get an odd number B = get a 5 or 6 What is P(A or B)? Alternative answer
- 21. Lots of possibilities then P(can get to work) = P(A clear or B clear or C clear) = 1 – P(A blocked and B blocked and C blocked
- 22. Problems with a device There are three common ways for a system to experience problems, with independent probabilities over a year A = overheats, P(A)=1/3 B = subcomponent malfunctions, P(B) = 1/3 C = damaged by operator, P(C) = 1/10 What is the probability that the system has one or more of these problems during the year? 1. 2. 3. 4. 5. 1/3 2/5 3/5 3/4 5/6 57% 24% 6% 4% 1 2 3 4 9% 5
- 23. Problems with a device There are three common ways for a system to experience problems, with independent probabilities over a year A = overheats, P(A)=1/3 B = subcomponent malfunctions, P(B) = 1/3 C = damaged by operator, P(C) = 1/10 What is the probability that the system has one or more of these problems during the year?
- 24. B A E.g. Throwing a fair dice, P(getting 4,5 or 6) = P(4)+P(5)+P(6) = 1/6+1/6+1/6=1/2 C
- 25. Rules of probability recap
- 26. Independent Events : Special Multiplication Rule If two events A and B are independent then P(A| B) = P(A) and P(B| A) = P(B): knowing that A has occurred does not affect the probability that B has occurred and vice versa. Probabilities for any number of independent events can be multiplied to get the joint probability.
- 27. Independent Events : Example E.g. A fair coin is tossed twice, what is the chance of getting a head and then a tail? P(H1 and T2) = P(H1)P(T2) = ½ x ½ = ¼. E.g. Items on a production line have 1/6 probability of being faulty. If you select three items one after another, what is the probability you have to pick three items to find the first faulty one?

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