L6.-Probability-of-Eventshahahahaha.pptx

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1. PROBABILITY OF EVENTS Learning Targets : 1. I can illustrate the probability of events , 2. I can find the probability of events of different types, and 3. I can solve problems involving probability.

2. PROBABILITY OF AN EVENT P(E) ⇒ is the ratio of the number of ways that an event can occur to the total number of possible outcomes when each outcome is equally likely to occur. P(E) = number of outcomes favorable to E total number of possible outcomes P(E) = number of elements in Event E number of elements in the Sample Space S = n(E) n(S) ≤ P(E) ≤ 1 0% ≤ P(E) ≤ 100%

3. A fair coin is tossed twice. Let A be the event of getting at least one head. Find, a . n(S) and n(A) , and b . the probability of getting at least one head . Example 1: Answers: a . S = { HH, TT, HT, TH}, n(S) = 4 A = { HT, TH, HH }, n(A) = 3 b . P(A) = n(A) n(S) = 3 4 or 75%

4. From 5 cards numbered 0, 1, 2, 3 and 4, two cards are drawn in succession without replacement, a . list the Sample space S , b . list the event that the sum of the numbers in the card is greater than 6 c . find the probability that the numbers will have a sum greater than 6. Example 2 :

5. From 5 cards numbered 0, 1, 2, 3 and 4, two cards are drawn in succession without replacement, a . list the Sample space S , Example 2: S = (0,1) (0,2) (0,3) (0,4) (1,2) (1,3) (1,4) (2,3) (2,4) (3,4) (1,0) (2,0) (3,0) (4,0) (2,1) (3,1) (4,1) (3,2) (4,2) (4,3) n(S) = 20

6. Example 2 : S = (0,1) (0,2) (0,3) (0,4) (1,2) (1,3) (1,4) (2,3) (2,4) (3,4) (1,0) (2,0) (3,0) (4,0) (2,1) (3,1) (4,1) (3,2) (4,2) (4,3) n(S) = 20 b . list the event that the sum of the numbers in the card is greater than 6 E = { (3,4) (4,3) } n(E) = 2

7. Example 2: S = (0,1) (0,2) (0,3) (0,4) (1,2) (1,3) (1,4) (2,3) (2,4) (3,4) (1,0) (2,0) (3,0) (4,0) (2,1) (3,1) (4,1) (3,2) (4,2) (4,3) n(S) = 20 c . E = { (3,4) (4,3) }, n(E) = 2, find the probability that the numbers will have a sum greater than 6 . P(E) = n(E) n(S) = 2 20 = 1 10 or 10% ∴ , the probability that the numbers will have a sum greater than 6 is 1/10 or 10% .

8. Find the probability of rolling a 4 on a die? an odd number? Example 3 : S = { 1, 2, 3, 4, 5, 6 }, n(S) = E(rolling a 4) = P(rolling a 4) = n(E) n(S) = 1 6 { 4 }, 6 n(E) = 1 E(rolling an odd) = S = { 1, 2, 3, 4, 5, 6 }, n(S) = 6 { 1, 3, 5 }, n(E) = 3 P(rolling a n odd ) = n(E) n(S) = 3 6 = 1 2

9. ⇒ For any event A, P(A) + P(A ‘ ) = 1 Probability of Complementary Events The probability that it will rain today is or P(A ‘ ) = 1 – P(A) Example 4 : 0.36. What is the probability that it will not rain? event A ⟹ it will rain event A ' ⟹ it will not rain P(rain) = 0.36 P(not rain) = 1 – 0.36 = 0.64 or 64 %

10. Probability of Complementary Events The probability of Sarah winning the P(A ‘ ) = 1 – P(A) Example 5 : election is 42%. What is the probability that she will lose? event A ⟹ she will win event A ' ⟹ she will lose P(winning) = 42% P(losing) = 100% - 42% = 58%

11. Formula for Odds If the probability that an event will occur is p (where p ≠ 0 and p ≠ 1), then Odds in favor of the event = p : ( 1 – p) Odds against the event = (1 – p) : p Note: Odds are usually in the form a : b, where a and b are integers with no common factor.

12. Find the odds in favor of and against rolling a sum of 6 with a pair of dice. Example 6: S = (1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6) n(S) = 36 E(sum of 6) = { (1,5) (2,4) (3,3) (4,2) (5,1) } n(E) = 5 P(sum of 6) = 5 36

13. Find the odds in favor of and against rolling a sum of 6 with a pair of dice. Example 6 : P(sum of 6) = 5 36 = p P(sum of ≠ 6) = 1 – 5 36 = 31 36 1 - p Odds in favor = p: (1 – p) = 5 36 : 31 36 = 5 : 31 ∴, odds in favor are 5 : 31 Odds against are 31 : 5

14. Formula for Odds Odds in favor of the event = p : ( 1 – p) Odds against the event = (1 – p) : p The chance of a rain tomorrow is 10%. What are the odds against raining? Example 7 : P(rain) = 10%, P(not rain) = 100% - 10% = 90% p 1 -p Odds against rain = (1 – p) : p = 90 : 10 = 9 : 1

15. Probability of Mutually Exclusive Events The probability of event A or event B that are mutually exclusive events P(A U B ) is equal to the sum of the probability of event A, P(A) , and the probability of event B, P(B) In symbols,   P(A ∪ B) = P(A) + P(B)

16. A die is rolled once. Find the probability that a 3 or a 5 will be rolled. Example 8 : P(3 ∪ 5) = P(3) + P(5) S = { 1, 2, 3, 4, 5, 6}, n(S) = 6 1 = 6 + 1 6 = 2 6 = 1 3 ∴, P(3 U 5) = 1 3

17. A block is chosen at random from a bag containing 6 white blocks, 4 black blocks and 12 red blocks. What is the probability that the block chosen will be a red block or a white block? Example 9 : P( R ∪ W ) = P( R ) + P(W) n(white) = 6 n(black) = 4 n(red) = 12 n(S) = 22 = 1 2 22 + 6 22 = 1 8 22 = 9 11 ∴, the probability of chosing a red or a white block is

18. Probability of Mutually Inclusive Events The probability of event A or event B that are inclusive is equal to the sum of the probabilities of A and B minus the probability of their common outcome or intersection. In symbols, P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

19. What is the probability of spinning an even number or a number less than 5 on a spinner numbered 1 – 8? Example 10 : S = { 1, 2, 3, 4, 5, 6 , 7, 8 }, n(S) = 8 P( E ∪ < 5 ) = P( E) + P(< 5 ) – P(E ∩ < 5) n( E ) = 4, n( < 5 ) = 4, n( E ∩ < 5 ) = 2 = 4 8 + 4 8 – 2 8 = 6 8 = 3 4 P( E ∪ < 5 ) Solution 1:

20. What is the probability of spinning an even number or a number less than 5 on a spinner numbered 1 – 8? Example 10 : S = { 1, 2, 3, 4, 5, 6 , 7, 8 }, n(S) = 8 P( E ∪ < 5 ) = n( E ∪ < 5 ) = = 6 6 n( E ∪ < 5 ) n(S) 8 = 3 4 P( E ∪ < 5 ) =   Solution 2:

21. Probability of Independent Events The probability of two independent events A and B is equal to the product of their individual probabilities, P(A) and P(B). In symbols , P(A ∩ B) = P(A) • P(B)

22. A coin is tossed twice. What is the probability of obtaining heads on the first toss and tails on the second toss? Example 1 1: P(H ∩ T) = P(H) • P(T) S = { H, T}, n(S) = 2 Single toss = 1 2 • 1 2 = 1 4 P(H ∩ T) S = HH, TT, HT, TH n(S) = 4 Two tosses P(H ∩ T) = 1 4

23. Probability of Dependent Events The probability of two dependent events A and B, with P(A) > 0 and P(B) > 0, is P(A ∩ B) = P(A) • P(B|A) P(B|A) means probability of B given that A has happened.

24. A box contains 40 cubes of the same sizes. Of these cubes, 17 are red, 13 are blue and the rest are yellow. If two cubes are drawn at random without replacement, a. what is the probability that the two yellow cubes will be drawn? b. what is the probability that a yellow cube is drawn on the first draw and blue cube on the second draw? Example 1 2:

25. a. what is the probability that the two yellow cubes will be drawn? Example 1 2: n( red cubes ) = 17 n( blue cubes ) = 13 n( yellow cubes ) = n( S ) = 40 10 2 cubes are drawn without replacement Find: P(Y and Y) P(Y ∩ Y) = P(Y) • P(Y | Y) = 10 40 • – 1 9 – 1 39 = 3 13 3 52 P(Y ∩ Y)

26. b. what is the probability that a yellow cube is drawn on the first draw and blue cube on the second draw? Example 1 2: n( red cubes ) = 17 n( blue cubes ) = 13 n( yellow cubes ) = 10 n( S ) = 40 2 cubes are drawn without replacement Find: P(Y and B ) P(Y ∩ B ) = P(Y) • P( B | Y) = 10 40 • 13 – 1 39 = 1 3 1 12 P( Y ∩ B )

27. Conditional Probability The conditional probability of an event refers to the probability of that event on the condition that another event has occurred. P(B|A) ⇒ represents the probability of B occurring on the condition that a first event A has already occurred ⇒ The sample space will be altered because some conditions are given.

28. a. What is the probability of a family with two children having two boys? b.Suppose you already know that the family described in (a) has at least one boy, now answer the question in (a). P(BB) = Example 1 3: a. S = { BB, GG, GB, BG}, n(S) = 4 n(BB) = 1 1 4 ∴, the probability of the family having 2 boys is

29. a. What is the probability of a family with two children having two boys? b.Suppose you already know that the family described in (a) has at least one boy, now answer the question in (a). P(BB) = Example 1 3: a. S = { BB, GG, GB, BG} n(S) = 4 n(BB) = 1 1 3 ∴, the probability of the family having 2 boys is 3

30. We roll a pair of dice once and see that the two numbers are not the same. Compute the probability that a. the sum is 7. b. the sum is 4. c. the sum is 12. Example 1 3:

31. We roll a pair of dice once and see that the two numbers are not the same. Compute the probability tha t (a) the sum is 7. Example 1 3: S = (1,1) (2,1) (3,1) (4,1) (5,1) (6,1) (1,2) (2,2) (3,2) (4,2) (5,2) (6,2) (1,3) (2,3) (3,3) (4,3) (5,3) (6,3) (1,4) (2,4) (3,4) (4,4) (5,4) (6,4) (1,5) (2,5) (3,5) (4,5) (5,5) (6,5) (1,6) (2,6) (3,6) (4,6) (5,6) (6,6) n(S) = 36 30

32. Compute the probability tha t (a) the sum is 7. Example 1 3: S = (1,1) (2,1) (3,1) (4,1) (5,1) (6,1) (1,2) (2,2) (3,2) (4,2) (5,2) (6,2) (1,3) (2,3) (3,3) (4,3) (5,3) (6,3) (1,4) (2,4) (3,4) (4,4) (5,4) (6,4) (1,5) (2,5) (3,5) (4,5) (5,5) (6,5) (1,6) (2,6) (3,6) (4,6) (5,6) (6,6) n(S) = 30 P(sum is 7) = 6 30 = 1 5 ∴, the prob. of the getting a sum of 7 is

33. Compute the probability tha t (b) the sum is 4 . Example 1 3: S = (1,1) (2,1) (3,1) (4,1) (5,1) (6,1) (1,2) (2,2) (3,2) (4,2) (5,2) (6,2) (1,3) (2,3) (3,3) (4,3) (5,3) (6,3) (1,4) (2,4) (3,4) (4,4) (5,4) (6,4) (1,5) (2,5) (3,5) (4,5) (5,5) (6,5) (1,6) (2,6) (3,6) (4,6) (5,6) (6,6) n(S) = 30 P(sum is 4) = 2 30 = 1 15 ∴, the prob. of the getting a sum of 4 is

34. Compute the probability tha t ( c ) the sum is 12 . Example 1 3: S = (1,1) (2,1) (3,1) (4,1) (5,1) (6,1) (1,2) (2,2) (3,2) (4,2) (5,2) (6,2) (1,3) (2,3) (3,3) (4,3) (5,3) (6,3) (1,4) (2,4) (3,4) (4,4) (5,4) (6,4) (1,5) (2,5) (3,5) (4,5) (5,5) (6,5) (1,6) (2,6) (3,6) (4,6) (5,6) (6,6) n(S) = 30 P(sum is 12) = 30 = ∴, the prob. of the getting a sum of 12 is

35. a. A card is drawn from a standard deck of 52 cards, find the probability of drawing a red card or a face card. b. 2 cards are drawn from a standard deck of 52 cards without replacement, what is the probability of drawing an ace on the first draw and another ace on the 2nd draw? c. 2 cards are drawn from a standard deck of 52 cards with replacement, what is the probability of drawing an ace on the first draw and another ace on the 2nd draw? Example 14 :

36. 52 Playing Cards

37. 13 clovers 13 spades 26 black cards 13 diamonds 13 hearts 26 red cards 52 cards ⟹ total

38. ⟹ 12 face cards 52 cards ⟹ 4 aces 3 Jacks 3 Queens 3 Kings ⟹ 36 number cards

39. a. A card is drawn from a standard deck of 52 cards, find the probability of drawing a red card or a face card. Example 14 : 26 red cards + 6 face cards P( RC or FC ) = 26 + 6 52 = 32 52 = 8 13

40. b. 2 cards are drawn from a standard deck of 52 cards without replacement, what is the probability of drawing an ace on the first draw and another ace on the 2nd draw? Example 14 : 4 aces P( A ) = 4 52 x and A 3 51 1 13 1 17 = 1 221

41. b. 2 cards are drawn from a standard deck of 52 cards with replacement, what is the probability of drawing an ace on the first draw and another ace on the 2nd draw? Example 14 : 4 aces P( A ) = 4 52 x and A 4 52 1 13 1 13 = 1 169

42. A committee of 4 is formed from 10 boys and 6 girls at random. Find the probability that there are 3 boys and 1 girl in the committee. Example 14 : P (3B and 1G) = number of outcomes favorable to the event Total number of possible outcomes P (3B and 1G) = C 4 16 C 3 10 x C 1 6 = 120 x 6 1820 = 720 1820 = 36 91

43. 1. In a class, there are 20 boys and 25 girls, if one student is chosen at random, what is the probability of a. choosing a boy? b. choosing a girl? 2. A coin is tossed 3 times. Find the probability of getting either all tails or getting a tail exactly once. 3. A die is rolled one time. What is the probability of getting an odd number or getting a number greater than 2? Exercises:

44. 4. A die is tossed twice. Find the probability of getting a 3 or a 4 on the 1st toss and a 1 or a 2 on the 2nd toss. 5. A card is drawn at random from a deck of 52 cards. Without replacing it, a 2nd card and a 3rd card is drawn. What is the probability that the 1st card drawn is an ace, the 2nd is a number card and the 3rd card is a black jack? Exercises:

45. 6. In a class in an international school, the mother tongue of students is either Chinese or English. The table below shows the corresponding numbers... A student is chosen at random. Find the probabilities that a. the mother tongue of the student is Chinese given that the student chosen is a boy. b. the student is a boy given that his mother tongue is Chinese. Exercises: Chinese English Boys 12 4 Girls 10 2

46. 7. A joint school function involves 4 different schools each school send a boy and a girl to the prepration meeting. After the meeting, they are asked to sit in a row for a photograph. Suppose the students are arranged at random, find the probability that all students of the same gender sit together. 8. In a lotto game of 6/42, 6 different numbers from 1 – 42 are chosen by the bettor. If all the 6 numbers are picked regardless of order, the bettor wins the jackpot prize. What is the probability of winning the jackpot? Exercises:

47. Thank you!!!