1. Probability (Kebarangkalian) EBB 341

2. Probability and Statistics Parameters population sample before observation after observation data random variables Statistical procedure inference probability statistics Sampling techniques

3. Definition of probability  Probability: Likelihood-kemungkinan Chance-peluang Tendency-kecenderungan Trend-gaya/arah aliran  P(A) = N A /N P(A) = probability of event A N A = number of successful outcomes of event A N = total number of possible outcomes

4. Example:  A part is selected at random from container of 50 parts that are known have 10 noncomforming units. The part is returned to container. After 90 trials, 16 noncomforming unit were recorded. What is the probability based on known outcomes and on experimental outcomes?  Known outcomes: P(A) = N A /N = 10/50 = 0.200  Experimental outcomes: P(A) = N A /N = 16/90 = 0.178

5. Probability theorems  Probability is expressed as a number between 0 and 1. (Theorem 1)  If P(A) is the probability that an event will occur, then the probability the event will not occur is 1.0 - P(A) or P(A) = 1.0 – P(A). (Theorem 2)  P(A) = probability of not event A

6. Example for Theorem 2:  If probability of finding and error on an income tax return is 0.04, what is the probability of finding an error-free or conforming return?  P(A) = 1.0 – P(A) = 1.0 -0.04 = 0.96

7. Probability theorems  For mutually exclusive events, the probability that either event A or B will occur is the sum of their respective probabilities P(A or B) = P(A) + P(B). (Theorem 3).  When events A and B are not mutually exclusive events, the probability that either event A or event B will occur is P(A or B or both) = P(A) + P(B) - P(both). (Theorem 4).  “mutually exclusive” means that accurrence of one event makes the other event impossible.

8. Example for Theorem 3:  What is the probability of selecting a random part produced by supplier X or by supplier Z? P(X or Z) = P(X) + P(Z) = 53/261 + 77/261 = 0.498  What is the probability of selecting a nonconforming part from supplier X or a conforming part from supplier Z? Supp lier No. Conform ing No. Noncon forming Total X50353 Y1256131 Z75277 TOTAL 25011261 P(nc X or co Z) = P(nc X) + P(co Z) =3/261 + 75/261 = 0.299

9. Example for Theorem 4:  What is the probability that a randomly selected part will be from supplier X or a nonconforming unit? P(X or nc or both) = P(X) + P(nc) –P(X and nc) = (53/261) + (11/261) – (3/261) = 0.234

10. Probability theorems  Sum of the probabilities of the events of a situation equals 1. P(A) + P(B) + … + P(N) = 1.0 (Theorem 5).  If A and B are independent events, then the probability that both A and B will occur is P(A and B) = P(A) x P(B) (Theorem 6).  If A and B are dependent events, the probability that both A and B will occur is P(A and B) = P(A) x P(B|A) (Theorem 7). P(B|A): probability of event B provided that even A has accurred.

11. Example for Theorem 6 & 7:  What is probability that 2 randomly selected parts will be from X and Y? Assume that the first part is returned to the box before the second part is selected (called with replacement). P(X and Y) = P(X) x P(Y) = (53/261) x (131/261) = 0.102  Assume that the first part was not returned to the box before the second part is selected. What is the probability? P(X and Y) = P(X) x P(Y|X) = (53/261) x (131/260) = 0.102 Since 1 st part was not returned, the was a total of 260.

12. Example: Theorem 7  What is the probability of choosing both parts from Z? P(Z and Z) = P(Z) x P(Z|Z) = (77/261) (76/260) = 0.086 Theorems 3 and 6-to solve many problems it is necessary to use several theorems:  What is the probability that 2 randomly selected parts (with replacement) will have one conforming from X and one conforming part from Y or Z? P[co X and (co Y or co Z) = P(co X) [P(co Y) + P(co Z)] = (50/261) [(125/261) + (75/261)] = 0.147

13. Counting of events  Many probability problems, such as those where the evens are uniform probability distribution, can be solved using counting techniques.  There are 3 counting techniques:  Simple multiplication  Permutations  Combinations

14. Simple multiplication  If event A can happen in any a ways and, after it has occurred, another event B can happen in b ways, the number of ways that both event can happen is ab.  Example.: A witness to a hit and run accident remembered the first 3 digits of the licence plate out of 5 and noted the fact that the last 2 were numerals. How many owners of automobiles would the police have to investigate? ab = (10)(10) = 100  If last 2 were letters, how many would need to be investigate? ab = (26)(26) = 676

15. Permutations  A permutation is the number of arrangements that n objects can have when r of them are used. For example:  The permutations of the word “cup” are cup, cpu, upc, ucp, puc & pcu  n = 3, and r = 3 = number of permutations of n objects taken r of them (the symbol is sometimes written as n P r ) n! is read “n factorial” = n(n-1)(n-2) …(1)

16. Example:  How many permutations are there of 5 objects taken 3 at a time? = 60  In the licence plate example, suppose the witness further remembers that the numerals were not the same

17. Combinations  If the way the objects are ordered in unimportant.  “cup” has 6 permutations when 3 objects are taken 3 at a time.  There is only one combinations, since the same 3 letters are in different order.

18. Formula  The formula for combination: where = number combinations of n object taken r at a time.

19. Discrete Probability Distributions Typical discrete probability distributions:  Hypergeometric  Binomial  Poisson

20. Discrete probability distributions  Hypergeometric - random samples from small lot sizes. Population must be finite Samples must be taken randomly without replacement  Binomial - categorizes “success” and “failure” trials  Poisson - quantifies the count of discrete events.

21. Hypergeometric  Occurs when the population is finite and random sample taken without replacement  The formula is constructed of 3 combinations (total combinations, nonconforming combinations, and conforming combinations):

22.  P(d) = prob of d nonconforming units in a sample of size n.  N = number of units in the lot (population)  n = number of unit in the sample.  D = number nonconforming in the lot  d = number nonconforming in the sample  N-D = number of conforming units in the lot  n-d = number of conforming units in the sample = Combinations of all units = combinations of nonconforming units = combinations of conforming units

23. Example  A lot of 9 thermostats located in a container has 3 nonconforming units. What is probability of drawing one nonconforming unit in a random sample of 4?  N = 9, D = 3, n = 4 and d = 1 Lot Sample nonconforming unit conforming unit

24.  Similarly, P(0) = 0.119, P(2) = 0.357, P(3) = 0.048.  P(4) is impossible- only 3 nc units.  The sum probability:  P(T) = P(0) + P(1) + P(2) + P(3) = 0.119 + 0.476 + 0.357 + 0.048 = 1.000

25. “or less” & “or more” probability  Some solutions require an “or less” or “or more” probability.  P(2 or less) = P(2) + P(1) + P(0)  P(2 or more) = P(T) – P(1 or less) = P(2) + P(3) + …

26. Binomial  This is applicable to the infinite number of items or have steady stream of items coming from a work center.  The binomial is applied to problem that have attributes, such as conforming or nonconforming, success or failure, pass or fail.  Binomial expansion:

27.  p = prob. an event such as a nonconform  q = 1-p = prob. nonevent such as conform  n = number of trials or the sample size  Since p = q, the distribution is symmetrical regardless of the value of n.  When p  q, the distribution is asymmetrical.  In quality work p is the proportion or fraction nonconforming and usually less than 0.15.

28. Binomial for single term  P(d)= prob. of d nonconforming  n = number of sample  d = number nonconforming in sample  p o = proportion(fraction) nc in the population  q o = proportion(fraction) conforming (1-p o ) in the population

29. Example  A random sample of 5 hinges is selected from steady stream of product, and proportion nc is 0.10.  What is the probability of 1 nc in the sample?  What is probability of 1 or less?  What is probability of 2 or more? q o = 1-p o = 1.00 - 0.10 = 0.90

30. P(1 or less) = P(0) + P(1) = 0.918 P(2 or more) = P(2) + P(3) + P(3) + P(4) = P(T) – P(1 or less) = 0.082

31. Poisson  The distribution is applicable to situations: that involve observations per unit time (eg. count of car arriving at toll in 1 min interval). That involve observations per unit amount (eg. count nonconformities in 1000 m 2 of cloth ).

32. Poisson  Formula for Poisson distribution  where c=count or number  np o =average count, or average number  e=2.718281

33. Poisson  Suppose that average count of cars that arrive a toll booth in a 1-min interval is 2, then calculations are:

34. Poisson  The probability of zero cars in any 1-min interval is 0.135.  The probability of one cars in any 1-min interval is 0.271.  The probability of two cars in any 1-min interval is 0.271.  The probability of three cars in any 1-min interval is 0.180.  The probability of four cars in any 1-min interval is 0.0.090.  The probability of five cars in any 1-min interval is 0.036.  …….