Download presentation

Presentation is loading. Please wait.

Published byEliezer Erwin Modified about 1 year ago

1
Bayes Rule for probability

2
Let A 1, A 2, …, A k denote a set of events such that An generalization of Bayes Rule for all i and j. Then

3
Example: We have three urns. Urn 1 contains 14 red balls and 12 black balls. Urn 2 contains 6 red balls and 20 black balls. Urn 3 contains 3 red balls and 23 black balls. An Urn is selected at random and a ball is selected from that urn. If the ball turns out to be red what is the probability that it came from the first urn? second urn? third Urn? Urn 1 Urn 2 Urn 3

4
Solution: Note: the desired conditional probability is in the reverse direction of the given conditional probabilities. This is the case when Bayes rule should be used Let A i = the event that we select urn i Let B = the event that we select a red ball

5
Bayes rule states

6
Example: Suppose that an electronic device is manufactured by a company. During a period of a week –15% of this product is manufactured on Monday, –23% on Tuesday, –26% on Wednesday, –24% on Thursday and –12% on Friday.

7
Also during a period of a week –5% of the product is manufactured on Monday is defective –3 % of the product is manufactured on Tuesday is defective, –1 % of the product is manufactured on Wednesday is defective, –2 % of the product is manufactured on Thursday is defective and –6 % of the product is manufactured on Friday is defective. If the electronic device manufactured by this plant turns out to be defective, what is the probability that is as manufactured on Monday, Tuesday, Wednesday, Thursday or Friday?

8
Solution: Let A 1 = the event that the product is manufactured on Monday A 2 = the event that the product is manufactured on Tuesday A 3 = the event that the product is manufactured on Wednesday A 4 = the event that the product is manufactured on Thursday A 5 = the event that the product is manufactured on Friday Let B = the event that the product is defective

9
Now P[A 1 ] = 0.15, P[A 2 ] = 0.23, P[A 3 ] = 0.26, P[A 4 ] = 0.24 and P[A 5 ] = 0.12 Also P[B|A 1 ] = 0.05, P[B|A 2 ] = 0.03, P[B|A 3 ] = 0.01, P[B|A 4 ] = 0.02 and P[B|A 5 ] = 0.06 We want to find P[A 1 |B], P[A 2 |B], P[A 3 |B], P[A 4 |B] and P[A 5 |B]. We will apply Bayes Rule

10
iP[Ai]P[Ai]P[B|Ai]P[B|Ai]P[Ai]P[B|Ai]P[Ai]P[B|Ai]P[Ai|B]P[Ai|B] Total

11
The sure thing principle and Simpson’s paradox

12
The sure thing principle Suppose Example – to illustrate Let A = the event that horse A wins the race. B = the event that horse B wins the race. C = the event that the track is dry = the event that the track is muddy

13
Proof:

14
Simpson’s Paradox Does Example to illustrate D = death due to lung cancer S = smoker C = lives in city, = lives in country

15
If we let Then the statement would be true using the Sure Thing Principle This logic is incorrect The events are not defined and do not make sense. The conditional probabilities are defined.

16
similarly Solution

17
is greater than whether depends also on the values of

18
than whether and

19
123 The Monty Hall Problem Behind one of the three doors there is a valuable prize. Behind the other two doors is a worthless prize. You are asked to pick one of the doors. After you have selected, Monty Hall opens one of the doors and reveals a worthless prize. He then asks you do you want to switch your choice.

20
1.Should you change your choice? 2.Should you keep your first choice? or 3.It does not matter. Solution Suppose you choice is door #1, and Monty reveals that door #3 has a worthless prize behind it. We can always renumber the doors so that this is the case. Let A i = the event that the valuable prize is behind door number i. i = 1, 2, 3. P [A 1 ] = P [A 2 ] = P [A 3 ] = 1 / 3 S = A 1 A 2 A 3 and A i A j =

21

22
Another Solution (the correct solution) The probability that you pick the correct door is 1 / 3. If you pick the correct door Monty will pick randomly between the two worthless doors. If you did not pick the correct door Monty will choose the worthless door to open with with probability 1. Let B i = the event that Monty opens door i. i = 1, 2, 3. Again P [A 1 ] = P [A 2 ] = P [A 3 ] = 1 / 3 and S = A 1 A 2 A 3 and A i A j =

23
Also We want to compute P [A 1 |B 3 ] andP [A 2 |B 3 ]. and

24

25
1 23 Another Problem We have three chests each having 2 drawers In chest 1 there is a gold coin in each drawer. In chest 2 there is a silver coin in each drawer. In chest 3 there is a gold coin in the top drawer and a silver coin in the bottom drawer..

26
One of the chests is selected at random. Then the drawer is selected at random. The coin in that drawer turns out to be gold. What is the probability that the coin in the other drawer is also gold? Is it ½ ? 1 23 Solution Let C i = the event that we select Chest i. i = 1, 2, 3. P [C 1 ] = P [C 2 ] = P [C 3 ] = 1 / 3 S = C 1 C 2 C 3 and C i C j =

27
We want to compute Let D 1 = the event that we select top drawer in the chest. G = the event the coin in the drawer is goldLet = (C 1 D 1 ) (C 1 D 2 ) (C 3 D 1 ) D 2 = the event that we select bottom drawer in the chest. P[C 1 |G].

28
Thus Comment: There are 6 drawers and three of those drawers contain gold coins. Of those three drawers two are in a chest that has a gold coin in the other drawer.

29
Random Variables an important concept in probability

30
A random variable, X, is a numerical quantity whose value is determined be a random experiment Examples 1.Two dice are rolled and X is the sum of the two upward faces. 2.A coin is tossed n = 3 times and X is the number of times that a head occurs. 3.We count the number of earthquakes, X, that occur in the San Francisco region from 2000 A. D, to 2050A. D. 4.Today the TSX composite index is 11,050.00, X is the value of the index in thirty days

31
Examples – R.V.’s - continued 5.A point is selected at random from a square whose sides are of length 1. X is the distance of the point from the lower left hand corner. 6.A chord is selected at random from a circle. X is the length of the chord. point X chord X

32
Definition – The probability function, p(x), of a random variable, X. For any random variable, X, and any real number, x, we define where {X = x} = the set of all outcomes (event) with X = x.

33
Definition – The cumulative distribution function, F(x), of a random variable, X. For any random variable, X, and any real number, x, we define where {X ≤ x} = the set of all outcomes (event) with X ≤ x.

34
(1,1) 2 (1,2) 3 (1,3) 4 (1,4) 5 (1,5) 6 (1,6) 7 (2,1) 3 (2,2) 4 (2,3) 5 (2,4) 6 (2,5) 7 (2,6) 8 (3,1) 4 (3,2) 5 (3,3) 6 (3,4) 7 (3,5) 8 (3,6) 9 (4,1) 5 (4,2) 6 (4,3) 7 (4,4) 8 (4,5) 9 (4,6) 10 (5,1) 6 (5,2) 7 (5,3) 8 (5,4) 9 (5,5) 10 (5,6) 11 (6,1) 7 (6,2) 8 (6,3) 9 (6,4) 10 (6,5) 11 (6,6) 12 Examples 1.Two dice are rolled and X is the sum of the two upward faces. S, sample space is shown below with the value of X for each outcome

35

36
Graph x p(x)p(x)

37
The cumulative distribution function, F(x) For any random variable, X, and any real number, x, we define where {X ≤ x} = the set of all outcomes (event) with X ≤ x. Note {X ≤ x} = if x < 2. Thus F(x) = 0. {X ≤ x} = {(1,1)} if 2 ≤ x < 3. Thus F(x) = 1/36 { X ≤ x} = {(1,1),(1,2),(1,2)} if 3 ≤ x < 4. Thus F(x) = 3/36

38
Continuing we find F(x) is a step function

39
2.A coin is tossed n = 3 times and X is the number of times that a head occurs. The sample Space S = {HHH (3), HHT (2), HTH (2), THH (2), HTT (1), THT (1), TTH (1), TTT (0)} for each outcome X is shown in brackets

40
Graph probability function p(x)p(x) x

41
Graph Cumulative distribution function F(x)F(x) x

42
Examples – R.V.’s - continued 5.A point is selected at random from a square whose sides are of length 1. X is the distance of the point from the lower left hand corner. 6.A chord is selected at random from a circle. X is the length of the chord. point X chord X

43
Examples – R.V.’s - continued 5.A point is selected at random from a square whose sides are of length 1. X is the distance of the point from the lower left hand corner. An event, E, is any subset of the square, S. P[E] = (area of E)/(Area of S) = area of E point X S E

44
Thus p(x) = 0 for all values of x. The probability function for this example is not very informative S The probability function

45
The Cumulative distribution function S

46
S

47

48

49
The probability density function, f(x), of a continuous random variable Suppose that X is a random variable. Let f(x) denote a function define for - < x < with the following properties: 1. f(x) ≥ 0 Then f(x) is called the probability density function of X. The random, X, is called continuous.

50
Probability density function, f(x)

51
Cumulative distribution function, F(x)

52
Thus if X is a continuous random variable with probability density function, f(x) then the cumulative distribution function of X is given by: Also because of the fundamental theorem of calculus.

53
Example A point is selected at random from a square whose sides are of length 1. X is the distance of the point from the lower left hand corner. point X

54
Now

55
Also

56
Now and

57
Finally

58
Graph of f(x)

Similar presentations

© 2016 SlidePlayer.com Inc.

All rights reserved.

Ads by Google