Combinatorics and probability

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1 Combinatorics and probability Maths 4 th ESO José Jaime Noguera 1

2 Organizing data: tree diagrams Draw the tree diagram for the problem: You have 3 seats and three people Andrea (A), Bob (B) and Carol (C). In how many ways can they be seated? _ B _ C ABC A _ C _ A ACA _ A _ C BAC B _ C _ A BCA _ A _ B CAB C _ B _ A CBA 6 WAYS 2

3 The fundamental counting principle If there are m possible results of event and n possible results of event B, then the amount of possible results of event A followed by event B are= m n Example: 1) In your wardrobe you have 3 trousers, 4 t-shirts and 2 pairs of shoes. How many different ways can you dress? 3 4 2=24 ways 2) In how many ways can 4 people be seated in 4 chairs? = 24 ways 3

4 Permutations FACTORIAL: 5!= =120 10!= = P n = n! It means: all the possible arrangements (or choices) of n elements when the order is important. Example: In how many ways can be seated 5 people? P 5 = 5! =

5 When we have to choose r elements out of n we have to use the permutation formula: n! P n, r = (n r)! n is all the possible choices r represents the items that we can choose. EXAMPLE: In how many ways can be seated 6 people in 4 chairs? n=6, r=4, then P 6,4 = 6! = 720 (6 4)! 2 =360 ways. 5

6 Permutations with repetition When repetition is allowed we have to use the formula PR n, r = n r Example: In how many ways can 7 workers be selected for 4 jobs in the following cases: a) A person can have just one job 7! P n, r = (7 4)! a) A person can have several jobs PR 7,4 = 7 4 6

7 IMPORTANT: in Spanish and valencià this formula is said VARIACIONES-VARIACIONS V n, r = n! (n r)! A good video about permutations 7

8 Combinations Now the order is not important. The number of combinations of n elements taken r at a time is: n! C n, r = n r! r! This number is also known as binomial coefficient (nombre combinatori-número combinatorio): n r = n! n r! r! 8

9 PROPERTIES n 0 = 1 n n = 1 n = n r n r n r + n r+1 = n+1 r+1 9

10 Random Experiments We roll a die. Then, the sample space is S={1,2,3,4,5,6,7} What type of Event is E={3} -> elementary event E={1,3,5} -> compound event (or event) E= number less than or equal to 6 -> sure event E={7} -> impossible event 10

11 EXAMPLES 1) We toss a coin. Find: a) Sample space: S={heads, tails} b) An elementary event: E={heads} or E={tails} c) An event: E={heads} or E={tails}} d) An impossible event: E={5} e) If E={heads}, find E = {tails} f) If E={tails}, find E = {heads} 11

12 2) We roll a die. Find: a) Sample space: S={1,2,3,4,5,6} b) An elementary event: E={2} c) An event : E={1,3} d) A sure event: E= natural number e) An impossible event: E= irrational number f) If E={1,3}, find E = {2,4,5,6} g) If E={4, 6, 1}, find E = {2,3,5} h) E= odd number : E={1,3,5} i) E= even number : E={2,4,6} j) E= prime number : E={2,3,5} 12

13 UNION: OR A B = elements that are in A OR in B EXAMPLE A={a, b, f, s} B={b, r, a} Then What elements are in A or in B? A B={a, b, f, s, r} 13

14 INTERSECTION: AND A B = elements that are in A AND in B EXAMPLE A={a, b, f, s, d, w, v} B={b, z, x, m, l, a, t} Then What elements are in A and B at the same time? A B={a, b} 14

15 Probability The probability (P) is a function that assigns to each event a number between 0 and 1 which represents the ease of occurrence of that event P( sure event)=1 P(impossible event)=0 If we toss a coin, what do you think is the P(heads)? 15

16 Laplace rule of succession If we assume that the events of our experiment have the same probability to occur (equiprobability) then: number of favorable outcomes P(E)= number of possible outcomes 16

17 Examples For tossing a coin: E = heads P E = 1 2 =0.5 For rolling a die: E = 5 P E = 1 = E = odd number P E = 3 = 1 = E = 2,5 P E = 2 = 1 = E = 7 P E = 0 =

18 Now, the question is: if I toss a coin 10 times, am I going to obtain exactly 5 heads and 5 tails? Let s do an experiment: Number of experiments n i Absolute frequency of obtain heads f i Relative absolute frequency h i = f i /n i

19 Law of large numbers Following the previous experiment: As the number of experiments increases, the relative absolute frequency of an event approaches to the theoretical probability (defined by Laplace rule of succession) That is known as the law of large numbers 19

20 Two good questions If we toss a coin 100 times Is it possible to obtain 100 heads? YES If we toss a coin 100 times Is it probable to obtain 100 heads? NO 20

21 Properties of probability P( sure event )=1 P( impossible event )=P( )=0 0 P E 1 P A = 1 P A If A B (compatible events) P A B = P A + P B P A B If A B = (incompatible events or mutually exclusive events) P A B = P A + P B 21

22 Examples We roll a die. A= prime number B= odd number Calculate P A B First of all. Are they compatible or incompatible? A={2,3,5} B={1, 3, 5} A B = 3,5, therefore, they are compatible because A B. Then, we have to use the formula P A B = P A + P B P A B = = 4 6 =

23 Conditional probability The conditional probability of an event (B) is the probability of that event (B) will occur if another event (A) has already occurred. We write this as P(B A) and we read it as probability of B given A. Very important: We say that two events are independent if the occurrence of one does not affect the probability of the other one. In another case, we say that they are dependent. 23

24 Examples We toss a coin twice: A = obtain HEADS in the first toss. B= obtain TAILS in the second toss. A and B are independent because obtaining tails in the second toss is not affected by the first toss. 24

25 We have a bag with 3 black balls and 5 white balls. We extract a ball and then we replace the ball in the bag A= take a white ball in the first extraction B = take a white ball in the second extraction A and B are independent because obtaining a ball in the second extraction is not affected by the first extraction (because we replace the ball after the first extraction) We extract a ball and then we do not replace the ball in the bag A= take a white ball in the first extraction B = take a white ball in the second extraction A and B are dependent because obtaining a ball in the second extraction is affected by the first extraction (because we do not replace the ball after the first extraction and then, in the second extraction, we have only 4 white balls, so the probability is different) 25

26 Conditional probability If A and B are independent, then P A B = P A P(B) If A and B are dependent, then P A B = P A P B A = P B P A B Moreover we can write P B A = P(A B) P(A) 26

27 Let s do an example We have a bag with 3 black balls and 5 white balls. We extract a ball and then we replace the ball in the bag A = take a white ball in the first extraction B = take a white ball in the second extraction They are independent, so: P A B = P A P B = =

28 We have a bag with 3 black balls and 5 white balls. We extract a ball and then we don t replace the ball in the bag A = take a white ball in the first extraction B = take a white ball in the second extraction They are dependent, so: P A B = P A P B A = = 0.34 Another example W = take a white ball in the first extraction B = take a black ball in the second extraction P W B = P W P B W = =

29 Tree diagrams The previous problems are ALWAYS easier to solve using tree diagrams. Let s do an example: A bag contains 3 black balls and 5 white balls. You pick a ball and you replace it back in the bag. Then you pick another ball from the bag. Calculate the probability to pick: i) two black balls ii) a white ball and after a black ball iii) a black ball in his second draw In these type of problems we put P(W,B) when we want to say : the probability of white ball in the first draw and black ball in the second draw. We really are calculating P W B where: W = white ball in the first draw B = black ball in the second draw 29

30 i) P B, B = = 0.14 ii) P W, B = = 0.23 iii) P(second ball black) = P(BB or WB) = P(B,B)+P(W,B) = =

31 Example You have a bag with seven blue balls and 3 red balls. You pick up a ball at random from the bag, but do not replace it and then pick again at random. Calculate the probabilities for the picks: (a) two red balls (b) no red balls (c) at least one blue ball (d) one ball of each color 31

32 a) P(R,R) = = 0.07 b) P(B,B) = = 0.47 c) P(at least one blue )= P(BB or BR or RB) = P(B,B) + P(B,R)+P(R,B) = = 0.93 Another option is P(at least one blue) = 1-P(no blue)=1-p(r,r)= = 0.93 d) P(BR or RB) = P(B,R) + P(R,B) = =

Lecture 3. Probability and elements of combinatorics

Introduction to theory of probability and statistics Lecture 3. Probability and elements of combinatorics prof. dr hab.inż. Katarzyna Zakrzewska Katedra Elektroniki, AGH e-mail: zak@agh.edu.pl http://home.agh.edu.pl/~zak