Section 2 Probability

1. Section 2 Probability William Christensen, Ph.D.

2. Definitions • Probability is the chance that some event or outcome will happen P - denotes a probability A, B, ...- denote specific events or outcomes P (A)- denotes the probability of event A occurring

3. Definitions • Probability is the chance that some event or outcome will happen • If there is NO CHANCE that an event will occur, then we can say the Probability of that event occurring is 0 (or 0%). • If an event is CERTAIN to occur, then we can say the Probability of that event occurring is 1 (or 100%).

4. 0  P(A)  1 Certain to occur Impossible to occur Definitions • So, we now know that the Probability of any event occurring must be between 0 and 1 (or 0% and 100%) • We can say this mathematically as follows (where A represents any event):

5. Certain 1 Likely 0.5 50-50 Chance Unlikely Impossible 0 Possible Probability Values (0 – 1)

6. How to Calculate Probability • Relative Frequency Approximation • Classical Approach • Subjective Probabilities

7. P(A) = number of times A occurred number of times trial was repeated How to Calculate Probability • Relative Frequency Approximation • This method requires that we observe or do an experiment and actually count the number of times event A occurs • Example: using Relative Frequency Approximation to test the probability of tossing a coin and getting ‘heads’, we would actually toss a coin a number of times and then calculate what percent of the time we got ‘heads’. Thus the formula here is: • So, for instance if we tossed a coin 100 times and got heads 52 times, then we would calculate the P(heads) = 52/100 = 0.52 or 52%.

8. s P(A) = number of ways A can occur = n number of different simple events How to Calculate Probability • Classical Approach • This is the method we will focus on in class • It works like this. If a procedure has n number of different simple events (e.g., if you toss a coin there are only two things that can happen, heads or tails, so in this case n=2), and each has an equal chance of occurring (e.g., you must have an equal chance of getting heads or getting tails), and there are s number of different ways that A can occur (e.g., if A represents getting heads then there is only one way of getting heads, so s=1), then:

9. s P(A) = number of ways A can occur = n number of different simple events How to Calculate Probability • Classical Approach • Using the Classical Approach we can therefore calculate the Probability of tossing a coin and getting head: P(heads) as, • s (number of ways heads can occur) =1 • n (number of possible outcomes is either heads or tails) n=2 • s / n = 1 / 2 = 0.50 or 50% • So P(heads) = 0.50 or 50% You must be able to use this method to calculate Probabilities, so practice, practice, practice

10. How to Calculate Probability • Subjective Probabilities • This is the guessing method • Using this method we simply guess or make an estimate of the Probability of some event (A) • Therefore, P(A) is whatever you estimate it to be • For example, I might ask you, “What do you think the chances are that it will rain tomorrow?” and you might guess something like 0.05 (5%). Thus P(rain) = 0.05 or 5%. • We probably all use this method a lot in real life, but it obviously is not very scientific or accurate

11. How to Calculate Probability Summary • Relative Frequency Approximation • Provides an approximation based on observation or experiment • The “Law of Large Numbers” states that the larger the number of observations, the nearer our Probability using this method will approach the true probability calculated using the Classical Approach • Example: The more times we toss a coin the closer our Relative Frequency Approximation will come to P(heads)=0.50 found in the Classical Approach • Classical Approach • Provides an actual probability and is the method of choice for class • Subjective Probabilities • Simply a guess or estimate

12. Example:Find the probability that a randomly selected person will be struck by lightning this year. The sample space consists of two simple events: the person is struck by lightning or is not. Because these simple events are not equally likely, we cannot use the Classical Approach and must use Relative Frequency Approximation or Subjective Probability. Using Relative Frequency Approximation we can research past events to determine that in a recent year 377 people were struck by lightning in the US, which has a population of about 274,037,295. Therefore, P(struck by lightning in a year)  377 / 274,037,295 1/727,000 or roughly 1 in a million

13. Example:On an ACT or SAT test, a typical multiple-choice question has 5 possible answers. If you make a random guess a question, what is the probability that your response is wrong? There are 5 possible outcomes or answers, and there are 4 ways to answer incorrectly. Random guessing implies that the outcomes in the sample space are equally likely, so we apply the Classical Approach to get: P(wrong answer) = 4/5 = 0.80 or 80%

14. Rounding Rule for Probabilities • Give the exact fraction or • Round off the final result to 3 significant digits • Example: round 0.3456789 to 0.346 • Example: round 0.000000158702 to 0.000000159

15. Complementary Events

16. P(A) (read “not A”) Complementary Events The complement of event A, denoted by A, consists of all outcomes in which event A does not occur.

17. Example: Testing CorvettesThe General Motors Corporation wants to conduct a test of a new model of Corvette. A pool of 50 drivers has been recruited, 20 or whom are men. When the first person is selected from this pool, what is the probability of not getting a male driver? Because 20 of the 50 subjects are men, it follows that 30 of the 50 subjects are women so, P(not selecting a man) = P(man) = P(woman) = 30 = 0.6 50

18. P(A) + P(A) = 1 Complementary Events • Another approach to finding P(A) (probability of not A) is to recognize that the P(A) (the probability of some event occurring) OR P(A) (the probability of some event not occurring) must total up to 1 (100% or certainty) • In other words: • This is a critical concept you must understand and remember

19. P(A) + P(A) = 1 P(A) = 1 - P(A) P(A) = 1 - P(A) Complementary Events • And, since • Then simple algebra tells us that: • and

20. Applying this rule to our previous Example: Testing CorvettesThe General Motors Corporation wants to conduct a test of a new model of Corvette. A pool of 50 drivers has been recruited, 20 or whom are men. When the first person is selected from this pool, what is the probability of not getting a male driver? P(not selecting a man) = P(man) = 1 – P(man) = 1 – (20/50) = 1 – 0.40 = 0.600 • Notice how we got the same answer, but in a different way. You must be able to do this type of problem.

21. Probability Addition Rule The “OR” Rule

22. Addition Rule • To find the probability that event A or event B occurs, we use the addition rule: P(A or B) = P(A) + P(B) where A and B are mutually exclusive or cannot occur at the same time • If A and B are not mutually exclusive (they can happen simultaneously) then: P(A or B) = P(A) + P(B) – P(A and B) where P(A and B) denotes the probability that A and B both occur at the same time

23. Addition Rule P(A or B) = P(A) + P(B) - P(A and B) we subtract P(A and B) to avoid double that area where both A and B occur P(A or B) = P(A) + P(B) Total Area = 1 Total Area = 1 P(A) P(B) P(A) P(B) Non-overlapping Events A and B are mutually exclusive Overlapping Events A and B NOT mutually exclusive

24. Men Women Boys Girls Totals Survived 332 318 29 27 706 Died 1360 104 35 18 1517 Total 1692 422 64 56 2223 Addition Rule Using a Contingency Table • The best way to approach a problem involving P(A or B) is to put the information into what’s called a Contingency Table (shown below). • Here is an example showing the passengers on the Titanic and whether they died or survived. We’ll use this information to work some examples.

25. Men Women Boys Girls Totals Survived 332 318 29 27 706 Died 1360 104 35 18 1517 Total 1692 422 64 56 2223 Titanic Example I • Let’s say we want to find the probability of randomly selecting a Man or Boy. In other words, find P(Man or Boy) • First, we need to know whether Man and Boy are mutually exclusive. Just think, is it possible to be both a Man and Boy at the same time – obviously not. We can also see this as we circle “Men” and “Boys” we see they do not intersect. • Thus, P(Man or Boy) = P(Man) + P(Boy) • Since there are 1692 men out of 2223 passengers, the P(Man) = 1692/2223 = 0.761 • There are 64 boys out of the 2223 passengers, so P(Boy) (the probability of randomly selecting a Boy) = 64/2223 = 0.029 • Finally, P(Man or Boy) = P(Man) + P(Boy) = 0.761 + 0.029 = 0.790

26. Men Women Boys Girls Totals Survived 332 318 29 27 706 Died 1360 104 35 18 1517 Total 1692 422 64 56 2223 Titanic Example II • Next, let’s see if we can find the probability of randomly selecting a Man or someone who Survived. In other words, find P(Man or Survived) • First, we need to know whether Man and Survived are mutually exclusive. Just think, is it possible to be both a Man and a Survivor at the same time – obviously YES. We can also see this as we circle “Men” and “Survived” we see they intersect or overlap. • Thus, P(Man or Survived) = P(Man) + P(Survived) – P(Man and Survived) • Since there are 1692 men out of 2223 passengers, the P(Man) = 1692/2223 = 0.761 • There are 706 Survivors out of the 2223 passengers, so P(Survived) = 706/2223 = 0.318 • There are 332 people who are both Men and Survived, so P(Man and Survived) = 332/2223 = 0.149 • Finally, P(Man or Survived) = P(Man) + P(Survived) - P(Man and Survived) = 0.761 + 0.318 – 0.149 = 0.930

27. Probability Multiplication Rule The “AND” Rule

28. Multiplication Rule • To find the probability that event A and event B occurs, we use the multiplication rule • How we approach this depends on whether or not event A and B are connected or depended on each other • IF event A and B are independent events (A does not effect B in any way), then: P(A and B) = P(A) x P(B) where A and B are independent events • IF event A and B are somehow dependent on each other, then: P(A and B) = P(A) x P(B/A) where A and B are dependent P(B/A) represents the probability of event B occurring after it is assumed that event A has already occurred (read B/A as “B given A”)

29. Example: Multiplication Rule and Independent Events (coin toss) • Let’s go back to our example of tossing a coin. We already determined that P(heads)=0.50 • Now let’s use the multiplication rule to find the probability of tossing a coin 3 times and getting heads every time. We state this as P(Head and Head and Head) • First, we have to determine whether these are independent or dependent events. Think about it, is there any influence or effect between coin tosses. I don’t think so. Just because you throw heads one time has absolutely no effect on what happens the next time you toss the coin. Every time you toss the coin it is “independent” of every other time you toss the coin

30. Example: Multiplication Rule and Independent Events (coin toss) • Thus, P(Head and Head and Head) = 0.50 x 0.50 x 0.50 = 0.125 or 12.5%. This is the probability of tossing a coin three times and getting heads all three times • You must know how to do these kinds of problems

31. Example: Multiplication Rule and Independent Events (babies) • Let’s try another problem. Let’s find the probability of a couple having 5 children and all five of them being girls. Find P(Girl and Girl and Girl and Girl and Girl) • We can safely say that having a boy or girl one time has no effect on the gender of the next child, so we have independent events. Let’s also assume that a couple has an equal chance of having a boy or a girl, so the P(Girl) = ½ = 0.500. • Thus, P(Girl and Girl and Girl and Girl and Girl ) = 0.50 x 0.50 x 0.50 x 0.50 x 0.50 = 0.0312 or 3.12%. Given our assumptions, this is the probability of having all girls among 5 children

32. Multiplication Rule • We’ve now worked a couple of examples in which we had independent events (coin toss and babies), using the formula: P(A and B) = P(A) x P(B) where A and B are independent events • Now let’s explore what happens when we have dependent events, using the formula: P(A and B) = P(A) x P(B/A) where A and B are dependent Note that we can also write this formula as: P(B/A) = P(A and B) / P(A)

33. Men Women Boys Girls Totals Survived 332 318 29 27 706 Died 1360 104 35 18 1517 Total 1692 422 64 56 2223 Multiplication Rule – Dependent Events • Let’s go back to our Titanic example and see if we can find the P(Man and Survived) using the multiplication rule. • First, is survival somehow effected or dependent on whether or not you are a man. Yes, I think so, since it was supposed to be “women and children first” in the lifeboats. So we have dependent events and must use the formula P(A and B) = P(A) x P(B/A), or in this case P(Man and Survived) = P(Man) x P(Survived / Man) • With 1692 of the 2223 passengers being Men, P(Man) = 1692 / 2223 = 0.761 • With 332 survivors among 1692 men, the probability you survived GIVEN you are a man P(Survived / Man) = 332 / 1692 = 0.196 • This is probably the step that students mess up on the most. Make sure you practice and understand this concept of P(B/A) and how to read it off a contingency table • Thus, P(Man and Survived) = P(Man) x P(Survived / Man) = 0.761 x 0.196 = 0.149 or 14.9% • You might notice how putting the data in a Contingency Table can help us simplify the problem. Notice how we can read right off the table that there are 332 people who are Men and Survived, out of 2223 passengers, so we can simply take 332 / 2223 = 0.149 or 14.9%

34. Multiplication Rule Test for Independence • I’ve given you examples of the multiplication rule with independent events and with dependent events. However, there may be some problems where you are not sure whether you are dealing with independent events or dependent events. In those situations, there is a simple test you can do. • First, it helps to put your data into a Contingency Table so you can clearly see how things are related • Next, check to see if P(B) = P(B/A) • If P(B) = P(B/A) then A and B are independent • If P(B) ≠ P(B/A) then A and B are dependent

35. Probability of “at least one”

36. Probability of “at least one” • “At least one” means one or more • Think about this, the “complement” of one or more is what? Isn’t it none. Thus: • Compliment of P(at least one) or P(at least one) = P(none) • Since we know that P(A) + P(A) = 1, then: • P(at least one) + P(none) = 1 since P(none) is equivalent to the compliment of P(at least one) • Finally, we can say P(at least one) = 1 – P(none) This is a critical concept that you must understand

37. Example: Probability of “at least one” • Using the formula we just learned: P(at least one girl) = 1 – P(no girls) Let’s find the probability of a couple having at least 1 girl among 3 children, or P(at least 1 Girl) among 3 children • To solve this problem the key is find P(no girls) among 3 children. So, if the couple had 3 children and none of them were girls, what did they have? Boys I hope! Thus P(no girls) is the same as P(Boy and Boy and Boy). Since we already know the P(Boy) = 0.50, the P(Boy and Boy and Boy) = 0.50 x 0.50 x 0.50 = 0.125 • Now we can solve the problem because we know P(no girls) = 0.125, so P(at least one girl) = 1 – P(no girls) = 1 – 0.125 = 0.875 or 87.5%

38. Counting Rule

39. Fundamental Counting Rule For a sequence of two events in which the first event can occur m ways and the second event can occur n ways, the events together can occur a total ofm • n ways. Example: if our two events are putting two letters of the alphabet together, and there are 26 letters or ways the first event can occur, and another 26 letters or ways the second event can occur, then there are 26 x 26 = 676 different ways that we can combine two letters.

40. Definition • The factorial symbol ! denotes the product of decreasing positive whole numbers. Thus: • n! = n (n-1) (n-2) (n-3) …… • Special exception: 0!= 1 • Example: 3! = 3 x 2 x 1 = 6 • In Excel, we use the formula =FACT(number) to calculate the factorial of any number

41. Factorial Rule A collection of ndifferent items can be arranged in n! different orders or ways. Example: A deck of 52 different cards can be arranged in 52! = 80,658,175,170,943,900,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000 different orders or ways.

42. Permutations Rule If we select r items from a total selection of ndifferent items, the number of permutations possible is n! / (n – r)! Example: If we want to know how many different 5 card selections there are from a deck of 52 different cards, we can use this formula. Thus, there would be 52!/(52-5)! = 311,875,200 different 5 card selections you could get from a deck of 52 cards. Note: permutations includes every possible ordering, thus every possible ordering of the same 5 cards adds to the possible selections when using this formula. Since, for example in poker, we really don’t care what order the cards are in as long as we have 5 particular cards in our hand, we should not use this Permutations Rule, but rather the Combinations Rule which we present next, in which we are not concerned with every possible ordering.

43. Combinations Rule If we select r items from a total selection of ndifferent items, the number of combinations possible is n! / (n – r)!r! Example: If we want to know how many different 5 card poker hands are possible from a deck of 52 different cards, we can use this formula. Thus, there would be 52!/(52-5)!5! = 2,598,960 different 5 card hands possible from a deck of 52 cards. Note: combinations does not care about the order in which we get the 5 cards, only what specific 5 cards we end up with.

44. Permutations vs. Combinations Which One to Use? When different orderings of the same items are to be counted separately, we have a permutation problem, but when different orderings are not to be counted separately, we have a combination problem.

45. Section 2 Probability E N D William Christensen, Ph.D.