1 MATH 109 Sampling without Replacement We now shall consider some probabilities that result when sampling without replacement either in order or without regard to order. The process of drawing cards illustrates the ordered sampling. . Th e cards 10 through Ace are considered to be ”High” cards. When Example 1 , what is the probability of drawing 5 cards in sequence without replacement (a) a Hig h card, then a Low card, then a Low card, then a High card, then a High card? (b) a Face card, then a Number card, then a Number card, then a Number card, then an Ace? (c) a Heart, then a Diamond, then a Diamond, then a club, then a Heart? (a) To start, there are 20 high cards and 32 low cards out of 52. By not Solution . replacing the cards before drawing the next, the numbers decrease by one as the next cards are drawn. Thus, the desired probability is (high, low, low, high, high) 20 32 31 19 18 ≈ . 0.0217564 × × × × 48 52 50 51 49 (b) To start, there are 12 face cards and 36 number cards out of 52. So the desired probability is (face, number, number, number, Ace) 36 4 35 34 12 0.0065934 ≈ . × × × × 48 49 51 52 50 (c) To start, there ar e 13 in each suit. Now the desired probability is H D D C H 13 13 13 12 12 . 0.001 ≈ × × × × 50 51 52 49 48 Sampling Without Regard to Order N Now suppose we have a population of elements that a re divided into two types: N B B A A elements, and Type II which has elements, where + Type I which has = . N For example, a standard deck of 52 playing cards can be divided in many ways. = A Type I could be “Hearts” and Type II could be “All Others.” Then there are = 13 € B = 39 Others. Hearts and € €

2 (52 cards) Total is A + B = N B objects of Type 2 (39 others) A objects of Type I (13 hearts) Sample n n Now suppose a random sample of size is taken all at once from the entire N population of sampling w objects. We then are ithout replacement and without regard to k order . We wish to find the probability of having exactly elements of Type I in this sample. € € € When sampling all at once without replacement, k of having exactl the probability elements of Type I is given by y C ( A , k ) × C ( B , n − k ) C , n ) ( N n The denominator comes from the total number of ways to choose from the N k A . The numerator comes from choosing from Type entire group of out of B I, then choosing the rest from of Type II. A aver The age number of objects of Type n is given by in the sample of size A , = × n p n N A A is the prob. of Type where = p N

3 Example 2 . Suppose a hand of 5 cards is dealt from a deck of 52. (a) What are the possibilities for the number of Hearts in the hand? (b) What is the probability of the re being exactly 2 Hearts in the hand? (c) What is the average number of Hearts that will be in the hand? (d) What is the average number of non - Hearts that will be in the hand? N B A . There are , = 52 cards = 39 others. We take a sample Solution = 13 Hearts, and n of size = 5. (a) We can have anywhere from 0 to 5 Hearts in the hand. (b) The probabi lity of exactly 2 Hearts is € € (13 , 2) × C ( 39 , C 3) 78 × 9139 712, 842 . = 0.27428 ≈ = C 5) (52, 960 2, 960 598, 2, 598, The denominator comes from the total number of ways to choose 5 cards from 52. The nu merator comes from choosing 2 out of 13 Hearts, then choosing 3 from the other 39 cards. A 13 (c) The average number of Hearts in a 5 card deal is . 1.25 = = × 5 n × 52 N (d) The average numb er of non - Hearts in a 5 card deal must be 5 – 1.25 = 3.75 . . Example 3 At a campus protest rally, there are 18 students and 12 faculty. The cops choose 20 people at random to bust for disorderly conduct. In this group of 20, (a) What are the possibil ities for the number of students busted? What are the possibilities for the number of faculty busted? (b) What is the probability of there being exactly 12 students busted? (c) What is the average number of students chosen in a sample of size 20? d) What is the average number of faculty chosen in a sample of size 20? ( N B A There are = 30 people, Solution. = 18 students, and ulty. We take a = 12 fac n sample of size = 20. (a) Since 20 are chosen and there are only 12 faculty, then at least 8 students must be € chosen. So anywhere from 8 to 18 students may be busted. If there are 8 students, then € there must be 12 faculty. If there are 18 students, then there must be only 2 faculty. So anywhere from 2 to 12 faculty would have to be busted. (b) The probability of exactly 12 students being busted is (18 , 12) × C ( 12, 8) C , 189 9, 18, 564 × 495 180 = . 0.305847 = ≈ C (30, 20) 015 30, 045, 30, 045, 015 The denominator comes from the total number of ways to choose 20 people from 30. The numerator comes from choosing 12 out of 18 students, then choosing 8 from 1 2 faculty.

4 18 (c, d) The average number of students in a sample of 20 is . And so the = 12 × 20 30 8 . average number of faculty in a sample of 20 is you. Of these Example 4 . There are 80 people serving jury duty. One of these people is 80 people, 14 are chosen at random to try a particular case. What is the probability that you are chosen for the case? B A Consider yourself the 1 person of Type . The Solution. . Then there are 79 of Type A being chosen (i.e., you) is probability of exactly 1 of Type C 1) × C ( , 13 ) (1 79, . 14 ) C ( 80, Since = 1, this value simplifies as follows: 1) C (1 , ! 79 ) ( 13 79, C 79 ! ! 14 14 13 ! × 66 ! = = . = × ! 80 (80 14 C ) , 13! 80 ! 80 ! 14 ! × 66 since there are 80 in the group and 14 So the chance of you being picked is 80 14 / are chosen. €

5 Exercises 1. When drawing 6 cards in order from a standard deck, what is the probability of drawing High card High card , , then a , then a a Low card , then a then a , then a Lo w card , Low card Low card (a) if drawing without repl acement of cards? - (b) if drawing with replacement and re shuffling of cards? 2. Suppose a hand of 7 cards is dealt. (a) What are the possibilities for the number of Aces in the hand? (b) What is the probability of there be ing exactly 1 Ace in the hand? (c) What is the average number of Aces in a 7 card deal? (d) What is the average number of non - Aces in a 7 card deal? 3. In a drama class, there are 12 females and 10 males. A group of 15 is to be chosen at random to read a screenplay. (a) What are the possibilities for the number of males chosen? For the number of females? (b) What is the probability of there being exactly 6 males chosen? is the average (c) What is the average number of males chosen in a sample of 15? What number of females chosen? 4. In another class, there are 10 males and 17 females. A random group of 14 is chosen. (a) What are the possibilities for the number of females chosen? For the number of males? (b) What is the probabilit y of there being exactly 9 females chosen? (c) What is the average number of females chosen in a sample of 14? What is the average number of males chosen? 5. You and your friend are in a police line - up of 20 people. A “witness” claims he can identi fy the 6 people who held up the bank. If the witness just fingers 6 people at random from the group of 20, what is the probability that both you and your friend are picked? (Simplify the factorials in fractional form.)

6 Solutions 32 20 29 30 19 31 0.0223736 1. (a) ≈ × × × × × 52 50 51 49 48 47 32 20 20 32 32 32 ≈ 0.021215 (b) × × × × × 52 52 52 52 52 52 B N A n = 52 cards, 2. There are = 48 others. The sample size is = 4 Aces, and = 7. (a) We can have anywhere from 0 to 4 Aces in the hand. (b) The probability of exactly 1 Ace is € € ( 1) C C 4, 48 , 6) ( × 048 086, 49, 4 × 12, 271 , 512 ≈ = . = 0.3669 (52, 7) C 784, , 560 133 133, 784, 560 The denominator comes from the total number of ways to choose 7 cards from 52. The numerator comes from choosing 1 out of 4 Aces, then choosing 6 from the other 48 cards. 4 7 0.53846 (c, d) The average number of Aces in a 7 card deal is ≈ , and the = × 7 13 52 7 . average number of non - Aces is ≈ 6.46154 7 − 13 N B A = 22 students, = 10 males, and = 12 females. We take a sample of 3. There are n size = 15. (a) Since there are only 12 females, we must have at least 3 males in the sample. But we € So the number of males must be from 3 to 10. can have at most 10 males in the sample. € If there are 3 males, then we must have 12 females. If there are 10 males, then we must have 5 females. So the number of females must be from 5 to 12. (b) The probability of exactly 6 males being ch osen is C (10, 6) × C ( 12, 9) 210 × 220 46, 200 0.2709. = = ≈ C ) (22, 15 544 544 170, 170, The denominator comes from the total number of ways to choose 15 students from 22. The numerator comes fro m choosing 6 out of 10 males, then choosing 9 from the 15 females. 10 , and the average ≈ (c) The average number of males in a sample of 15 is 6.818 × 15 22 12 . ≈ 8.182 number of females is × 15 22

7 N B A = 27 students, = 10 males. We take a sample of = 17 females, and 4. There are n size = 14. (a) Since there are only 10 males, we must have at least 4 females in the sample. The € number of females must be from 4 to 14. And the number of males must be from 0 to € 10. (b) The probability of exactly 9 females being chosen is (17 C 9) × C ( 10, 5) , 120 126, 6, 252 × 24, 310 = . 0.3054 = ≈ C ( 27 , 14) 058, 300 20, 058, 300 20, The denominator comes from the total number of ways to choose 14 students from choosing 5 from the 27. The numerator comes from choosing 9 out of 17 females, then 10 males. 17 ≈ 8.8148 , and the (c) The average number of females in a sample of 14 is × 14 27 10 ≈ average number of males is 5.1852 . × 14 27 B A 5. You and your friend are the 2 persons of Type . Then there are 18 of Type . The A being chosen (i.e., both you and your friend) is probability of exactly 2 of Type 2, ) × C (18, 4 ) ( 2 C . 6 ) 20, C ( Since = 1, this value simplifies as follows: ) 2, ( 2 C 18 ! 6 5 × 18! C (18, 4) 5 6 ! 6 4 ! × ! 14 = × = = = × ! 20 ! ) 6 4 20 × 19 20 ( 20 ! C 19 20, 6 ! 14 ! × 6 , then the chance of your friend In other words, the chance of you being picked is 20 3 5 5 6 . The chance of both of you being picked is = being picked is ≈ 0.078947 . × 19 20 19 38