Basic Probability Models. Ping-Shou Zhong

Transcription

1 asic Probability Models Ping-Shou Zhong 1

2 Deterministic model n experiment that results in the same outcome for a given set of conditions Examples: law of gravity 2

3 Probabilistic model The outcome of the experiment can not be predicted with certainty. Examples: a: toss a coin b: number of people waiting in a bus stop c: drawing two cards from a set of 52 cards 3

4 asic probability concepts Sample spaces Events Probability measure 4

5 Sample space Definition: collection of all possible outcomes. sample space is discrete if it has finite or countable infinite number of outcomes. Example 1: Toss a coin (finite discrete) The sample space is S={H,T}, where H: head and T: tail Example 2: Number of people in a store (countable infinite discrete). S={0,1,2,3, } Example 3: Lifetime of a bulb: S={t: 0 t < } (continuous) 5

6 Example 4: toss two coins. Two possible sample spaces: S={HH,TH, HT, TT} each outcome being an ordered sequence of results; S={0,1,2} each outcome being a possible number of heads obtained. 6

7 Example 5: rolling a pair of dice The sample space can be: 7

8 The sample space could also be: 8

9 Events n event is a subset of a sample space. n event is said to happen if the outcome of an experiment is a member of. Example: rolling a die ={3}, ={3,4}, C={1,2,3} If the outcome is 3, we will say, and C happen. 9

10 ={at least one with 6 points} 10

11 Set operations: events combination x x c and event of }:The complement : { happens or either : happen and both : happens happens : 11

12 Properties of set operations c c c c c c d C C C C c C C C C b a ) (, ) ( Law : DeMorgan's ) ( ) ( ) ( ), ( ) ( ) ( Distributive Laws : ) ( ) (, ) ( ) ( ssociativity : :, Commutativity : : 12

13 Venn diagram: ( ) c = c c 13

14 Probability measure set function P defined on a set collection F that satisfies the following axioms (a) P() 0 for all F (b) P(S)=1 (c) If 1, 2, F are pairwise disjoint, then P i=1 i = i=1 P( i ) The triple S, F, P is called a probability space 14

15 Some properties of P 1. P = 0, P S = P 1 for F 3. P c = 1 P 4. If, then P P 5. P + P c = P 6. P = P + P P 7. P 1 2 n = n i=1 P i 1 i<j n P( i j ) n+1 n P( i=1 i ) 15

16 Discrete probability measure For a discrete sample space S = ω i with P({ω i }) 0 and P(S)=1. Define a measure P = ω i P({ω i }) for any F. Probability measure of equally likely outcomes. If P ω i = 1 where N(s) is the total number of outcomes, then N(S) Number of sample points in P = N() N(S) Total number of sample points in S 16

17 Example 1 Throwing a pair of dice. What is the probability of ={ at least one 6 }? Sample space (treat the dice distinguishable): S 1 ={11,12,13,14,15,16,21,22,23,24,25,26,31,32,33,34,35,36 41,42,43,44,45,46,51,52,53,54,55,56,61,62,63,64,65,66} Total number of sample points in the sample space are: N(S 1 ) = 6 6 = 36 Method 1: Listing all the outcomes in event ={16,26,36,46,56,66,61,62,63,64,65} N() = 11 P = 11/36 17

18 Question Why we need to treat the dice distinguishable? If not, the sample space is S 2 ={11, 22, 33, 44, 55,66, 12,13,14,15,16, 23,24,25,26, 34,35,36, 45,46, 56} ={16,26,36,46,56,66} P = N() N(S 2 ) = (Why?) Which one is right? 18

19 Reason For example {12} in sample space S 2 representing {12, 21} in sample space S 1 ccording to the sampling scheme, each sample point in S 1 is equal likely. ut this is not for S 2. P({12})= 2 36 in sample space S 2. 19

20 Method 2: Use the complement and multiply rule. ecause P = 1 P c, we calculate P c. c ={No 6 appears} N( c ) = 5 5 = 25 P c = Hence, P = 1 P c =11/36 20

21 Proportion of throws with at least one Proportion of throws with at least one Computer simulations Number of throws Number of throws 21

22 Example 2: Poker game 52-card deck containing 13 ranks and each rank has 4 suits: Clubs, Spades, Hearts and Diamonds. 22

23 Categories Straight flush: 5 consecutive cards with the same suit Four of a kind: 4 cards are of one rank Full house: 3 cards of one rank and two of another Flush: 5 cards with the same suit Straight: 5 consecutive cards lways classified in the higher category if it satisfies the definition ces is counted as either low or high 23

24 Sample space Let D be the set of 52 cards Sample space S = {: D and N = 5} N S = 52 5 = =2,598,960 5! (Without replacement, unordered sample) 24

25 Probability of full house Count the number of possible full house Choosing ranks: firstly for rank with 3 cards, then for rank with 2 cards Number of ways=13 12 Note that the second rank can not be the same as the first one. Otherwise it will be classified as Four-of-a-kind Choosing suits: Number of ways= 4 3 Total number of possible full house = =3744 P(Full House)=3744/ =1/

26 Probability of flush Choosing suits: 4 ways Choosing ranks: = 1277ways straight flush Total number of flush =4 1277=5108 P(flush)=5108/2,598,960=1/508 26

27 Example 3: Matching Problems n couples attend a dance party. The partners of the women are randomly assigned among the men. What is the probability that ={at least one woman dancing with her husband}? Let us label n couples by 1, 2, 3,, n and denote E i ={ Woman i dances with her husband}. = n i=1 E i 27

28 pply inclusion-exclusion principle n n P = P E i = i=1 P E i 1 i<j n P(E i E j ) n+1 P(E 1 E 2 E n ) i=1 P E i = n 1! n! P E i E j = n 2! n! P E 1 E 2 E 3 E r = n r! n! Number of summands n n 2 n r 28

29 P n i=1 E i = n n 1! n! n 2 n 2! n! + n n 3! + 3 n! = 1 1 2! + 1 3! + 1 n+1 1 n! = e 1 when n is large This method can also be applied to calculate the probability of exactly k (k n) letters and envelops matching each other. 29

30 Example 4: Coupon Problem company randomly places one of a set of 4 cards into each packet of its product. nyone who collect a complete set of 4 cards can claim a prize. What is the probability that you will able to ={claim at least one prize if you buy 10 packets}? What is the complement event to? c ={can not claim prize} ={at least one card in the set can not be found in 10 packets} 30

31 Define E i ={can not find card i (i=1,2,3,4) in 10 packets} c = 4 i=1 E i P = 1 P c 4 = 1 P( i=1 E i ) 4 = 1 i=1 P E i + i<j P E i E j i<j<k P E i E j E k + P(E 1 E 2 E 3 E 4 ) 31

32 P E i =( 3 4 )10 P E i E j =( 2 4 )10 P E i E j E k =( 1 4 )10 P(E 1 E 2 E 3 E 4 )=0 P =1-4 ( 3 4 )10 +6 ( 2 4 )10 4 ( 1 4 )10 =