1. Chance We will base on the frequency theory to study chances (or probability).

2. Definition The chance of something gives the percentage of time it is expected to happen, when the basic process is done over and over again, independently and under the same conditions.

3. Examples One simple game of chance involves betting on the toss of a coin. The process of tossing the coin can be repeated over and over again, independently and under the same conditions. The chance of getting heads is 50%: in the long run, heads will turn up about 50% of the time.

4. Examples A die is a cube with 6 faces, labelled: 1, 2, 3, 4, 5, 6. When we roll a die, the faces are equally likely to turn up. The chance of getting a 1 is 1 in 6, i.e. 16.7%. This means that if the die is rolled over and over again, repeating the basic chance process under the same conditions, in the long run a 1 will show about 16.7% of the time.

5. Properties of chances If something is impossible, it happens 0% of the time. If something is sure to happen, then it happens 100% of the time. Chances are between 0% and 100%. The chance of something equals 100% minus the chance of the opposite thing. For example, suppose we are playing a game, and have a 45% chance to win. This means we expect to win about 45% of the time, and we also must expect to lose the other 55% of the time.

6. Compute the chances Based on the frequency theory, the chance of some event is the ratio: The frequency of some event / the total number of the events.

7. Example A box contains red marbles and blue marbles. One marble is drawn at random from the box (each marble has an equal chance to be drawn). If it is red, you win $1. If it is blue, you win nothing. You can choose between two boxes: Box A contains 3 red marbles and 2 blue ones; Box B contains 30 red marbles and 20 blue ones. Which box offers a better chance of winning, or are they the same?

8. Solution The two boxes offer the same chance of winning. Based on the frequency theory, in the long run, in box A, each of the 5 marbles will appear about 1 time in 5. So the red marbles will turn up about 3/5 of the time. The “3” is the “frequency” of red marbles, and the “5” is the “total” number of the events. So the chance is 60%.

9. Solution Similarly, in box B, each of the 50 marbles will turn up about 1 time in 50. So the red marbles will turn up about 30/50 of the time, your chance of winning is again 60%. So by comparing the ratios, we know that the chance of winning is the same.

10. Two main models 1. Draw two tickets at random with replacement from the box: 1, 2, 3, … (Suppose in the box, we mark the tickets as 1, 2, 3, …) This asks you to do the following process: shake the box, draw out one ticket at random (equal chance for all tickets), mark down the number, put it back in the box, shake again, draw a second ticket at random, mark down the number, and put it back in the box.

11. Two main models 2. Draw two tickets at random without replacement from the box: 1, 2, 3, … This asks you to do the following process: shake the box, draw out one ticket at random, set it aside, draw out a second ticket at random.

12. Remark When you draw at random, all the tickets in the box have the same chance to be picked. The two models can be applied to more general cases: e.g. (1) draw tickets at random from the box with more tickets---- 30, 50, 100 or more tickets; (2) draw more tickets at random----draw 5, 10, 100 or more tickets from the box.

13. The multiplication rule The chance that two things will both happen equals the chance that the first will happen, multiplied by the chance that the second will happen given the first has happened.

14. Example A box has three tickets, colored red, white and blue: R, W, B. Two tickets will be drawn at random without replacement. What is the chance of drawing the red ticket and then the white?

15. Solution Since we draw tickets at random, each of the ticket has the same chance to be picked. So on the first draw, the chance of drawing R and left with W and B in the box is 1/3. On the second draw, the chance of drawing W is ½, since there are only 2 tickets remaining in the box. So the chance of drawing R and then W is 1/3 x ½ = 1/6, or 16.7%.

16. Remarks If we exhaust all the possibilities of drawing 2 tickets from the box: R & W, R & B, W & B, W & R, B & R, B & W. We see that the chance is 1/6. This coincides the result that we got.

17. Remarks We have to be cautious to use the arguments in the above solution. Because the chance or probability means the percentage of time it is expected to happen. It represents the percentage in the long run, not just one time. A better understanding could be: suppose you start with 600 people, each of these people holds a box: R, W, B. About 200 of them will get R on the first draw. Of these 200 people, about 100 will get W on the second. So the chance is 100/600 = 1/6 = 16.7%.

18. More examples Two cards will be dealt off the top of a well- shuffled deck. What is the chance that the first card will be the seven of clubs and the second card will be the queen of hearts? Note. A deck of cards has 4 suits: clubs, diamonds, hearts, spades. There are 13 cards in each suit: 2 through 10, jack, queen, king, ace. So there are 4 x 13 = 52 cards in total.

19. More examples Solution. The chance that the first card will be the seven of clubs is 1/52. Given the first card was the seven of clubs, there is 51 cards remaining, so the chance that the second card will be the queen of hearts is 1/51. Therefore, the chance of getting both cards is 1/52 x 1/51 = 1/2652 ≈ 0.04%.

20. More examples A deck of cards is shuffled, and two cards are dealt. What is the chance that both are aces? Solution. The chance that the first card is ace equals 4/52. Given the first card is an ace, there are 3 aces in the 51 remaining cards. So the chance of a second ace equals 3/51. So the chance that both are aces is 4/52 x 3/51 = 12/2652 ≈ 0.45%.

21. More examples A coin is tossed twice. What is the chance of a head followed by a tail? Solution. The chance of a head on the first toss equals ½. No matter how the first toss turns out, the chance of tails on the second toss equals ½. So the chance of heads followed by a tail is ½ x ½ = ¼.

22. Conditional probabilities The chance of the event given something has already happened.

23. Example A deck of cards is shuffled and the top two cards are put on a table, face down. You win $1 if the second card is the queen of hearts. (a) What is your chance of winning the dollar? (b) You turn over the first card. It is the seven of clubs. Now what is your chance of winning?

24. Solution (a) Note that the bet can be settled without even looking at the first card. The second card is all you need to know. When the deck is shuffled, this brings the cards into random order. The queen of hearts has to be somewhere. There are 52 possible positions. So there is 1 chance in 52 for this card to be the second. Hence the chance of winning is 1/52.

25. Solution (b) Now the first card is already turned over. This has already happened. There are 51 cards left, and they are in random order. So event that the queen of hearts is the second one has the chance 1 in 51. Hence the chance of winning is 1/51.

26. Remark In part (b), 1/51 is a conditional probability (or conditional chance). The problem puts a condition on the first card. The conditional probability reads the second card is the queen of hearts given the first card is the seven of clubs. (Format: event A given event B, as we already seen on previous examples.) In part (a), 1/52 is an unconditional probability (or unconditional chance). There is no condition on it.

27. Formula for conditional probability Suppose we have two events: event A and event B. We want to compute the conditional probability for event A given event B. Then the formula for this will be: P(A|B) = P(A, B) / P(B). In fact, this is a reformulation of the multiplication rule: P(A, B) = P(A|B) x P(B).

28. Formula for conditional probability For example, in part (b) of our previous problem, the conditional probability can be expressed as: P(2 nd card is queen of hearts | 1 st card is seven of clubs). In part (a), the unconditional probability reads: P(2 nd card is queen of hearts). Note: Sometimes, it is easy to figure out P(A|B) as part (b) in the problem, so we may apply the multiplication rule to calculate P(A, B). (See the previous examples.) Sometimes, it will be easy to find P(A, B), then we could apply the formula for conditional probability to compute P(A|B).

29. Formula for conditional probability By using the formula, we may obtain the result in part (a) in an alternative way: P(2 nd card is queen of hearts) = P(2 nd card is queen of hearts | 1 st card is others) x P(1 st card is others) = 1/51 x 51/52 = 1/52. There are 51 chances in 52 when 1 st card is others. This is where 51/52 comes from. Given 1 st card is others, the conditional probability is 1/51. Hence we obtain the result.

30. Independence

31. Definition Two events are independent if the chances (or probabilities) for the second given the first are the same, no matter how the first one turns out. Using an equation, we can express: P(A|B) = P(A), or equivalently: P(A, B) = P(A) x P(B). Otherwise, the two events are dependent.

32. Example Someone is going to toss a coin twice. If the coin lands heads on the second toss, you win a dollar. (a) If the first toss is heads, what is your chance of winning the dollar? (b) If the first toss is tails, what is your chance of winning the dollar? (c) Are the tosses independent?

33. Solution (a) If the first toss is heads, there is a 50% chance to get heads the second time. (b) If the first toss is tails, the chance is still 50%. (c) No matter how the first toss turns out, the chances for the second toss stay the same: 50%. So they are independent.

34. Example Two draws will be made at random with replacement from a box of tickets: 1,1,2,2,3 (tickets marked by numbers). (a) Suppose the first draw is 1. What is the chance of getting a 2 on the second draw? (b) Suppose the first draw is 2. What is the chance of getting a 2 on the second draw? (c) Are the draws independent?

35. Solution Since we draw tickets with replacement, the second draw is always made from the same box with 1,1,2,2,3. So no matter what we draw on the first, the chance of getting 2 on the second stays the same: two in five, i.e. 2/5 = 40%. This shows the draws are independent.

36. Example If we modify the previous example a little bit: the draws are made without replacement. Are the answers stay the same?

37. Solution (a) If the first draw is 1, then before the second draw, the box has tickets: 1,2,2,3. So the chance for drawing 2 becomes 50%. (b) If the first draw is 2, then before the second draw, the box has tickets: 1,1,2,3. So the chance for drawing 2 becomes 25%. (c) By above arguments, the draws are dependent.

38. Remark When drawing at random with replacement, the draws are independent. Without replacement, the draws are dependent. (Exclude some extreme cases.) When we have a large number of tickets (say 10,000), we can treat the draws are independent even though we draw tickets without replacement. (The probabilities only change a little.) When the numbers on the tickets are identical (say identically 1), then draws are independent even though we draw tickets without replacement.

39. Apply the formula Example: A box has three tickets, colored red, white, and blue. (Mark as R,W,B.) Two tickets will be drawn at random with replacement. What is the chance of drawing the red ticket and then the white?

40. Apply the formula Solution. Recall that if two events are independent, then: P(A, B) = P(A) x P(B). Since we draw tickets with replacement, they are independent. So the probability is 1/3 x 1/3 = 1/9.

41. Remark If we compare this example to the example on slide 14, we find that it is easier this time. We don’t need to work with the conditional probabilities. This is how independence matters. The independence formula is a special case of the multiplication rule.

42. Summary The probability / chance of something gives the percentage of times the thing is expected to happen, when the basic process is repeated over and over again. Probabilities are between 0% and 100%. Impossibility is represented by 0%, certainty by 100%. The chance of something equals 100% minus the chance of the opposite thing.

43. Summary The multiplication rule: P(A, B) = P(A|B) x P(B). Two events are independent if the chances for the second one stay the same no matter how the first one turns out: P(A|B) = P(A). Consequence of independence: P(A, B) = P(A) x P(B).

44. Summary When we draw tickets at random, all tickets in the box share the same chance to be picked. Draws made at random with replacement are independent. Without replacement, the draws are dependent.