The following story, taken from the book by Polya, Patterns of Plausible Inference, Vol. II, Princeton Univ. Press, 1954, p.101, is also quoted in the book by Szekely, Classical paradoxes of probability theory, D. Reidel Publishing Company, 1986. D. Tel shook his head as he finished examining his patient. You have a very serious disease, he said, of ten people who have got this disease only one survives. As the patient was sufficiently scared by this information, D. Tel began to console him: But you are very lucky sir, because you came to me. I have already had nine patients who all died of it, so you will survive. Terminology: n(e)=the number of ways an event can occur with success. n(s)=the number of ways an event can occur, assuming all ways are equally likely to occur. The probability of an event E, n(s) assuming all events are equally likely to occur. Example: Suppose you roll a die. It can land with a 1,2,3,4,5, or 6 showing. n(s)=6. Question: What is the probability of getting a 1 or a 2 showing? A success is if it lands 1 or 2. So n(e)=2, and n(s) = 2 6 = 1 3. Odds: The odds of an event occurring, s(e) = n(e) : n(e ), where n(e )=the number of ways of having a failure and n(e) is the number of ways of having a success.. Example: to find the odds of getting a 1 or 2 with a single roll of a die :
n(e) = the number of ways of having a success = 2. n(e )=the number of ways of having a failure. You have a failure if it lands 3,4,5, or 6. So n(e )=4. Therefore s(e) = n(e): n(e ) = 2 : 4 = 1 : 2 What is the probability of getting (a total of) 6 when a pair of dice is rolled? Wrong way: It is possible to get a total of 2,3,4,5,6,7,8,9,10,11, or 12. So the probability of getting a 6 is n(e)/n(s)=1/11 since there is one way to get a sum of 6, hence n(e)=1, and there are 11 possible sums so n(s)=11. The last sentence contains an error. What is the error in the above calculation? Not all ways are getting a sum are equally likely. For example, there is only one way to get a sum of two: Both dice must show one. However, if you consider the two dice as die 1 and die 2, you could have die 1 = 1 and die 2 = 5 or die 1 = 2 and die 2 = 4, etc. We must count the ways so all outcomes are equally likely. There are 6 ways for die 1 to land and 6 ways for die 2 to land, and counting this way there are 36 ways both dice can land. so n(s)=36. Die 1 and die 2 can land as (1,5), (2,4), (3,3), (4,2), (5,1) so n(e)=5. Therefore p(e) = 5/36 A coin is tossed ten times. What is the probability that it lands heads all ten times? There are two ways the coin can land on the first toss {heads, tails }. There are two ways the coin can land on the second toss. So there are 2 2 ways the coin can land on the first two tosses. Continuing, there are 2 2 2 2 2 2 2 2 2 2 = 2 10 = 1024 ways the coin can land in ten tosses. n(e)=the number of ways it can land heads all ten times=1.
/n(s) = 1/1024. A random elevator rider is equally likely to exit on one of six floors. Three random riders enter the elevator. What is the probability that they all exit on different floors? Let n(e) be the number of ways that they can exit on different floors. The first rider can exit on any one of 6 floors. After that, the second rider has 5 choices, and after that the third rider has 4 choices. So n(e) = 6 5 4. The total number of ways three riders can exit is n(s) = 6 6 6. n(s) = 6 5 4 6 6 6 = 5 9 Three people are chosen at random. What is the probability that they all have different birthdays? What is the probability that at least two of the three people have the same birthday? Assume a year has 365 days (so we are ignoring the complication of leap years.) n(s)=the number of ways three people can have birthdays = 365 3. Let n(e)=the number of ways three people can have different birthdays. n(e) = 365 364 363 = 365 P 3. So n(s) = 365 364 363 365 365 365.992 Let E be the event that at least two people have the same birthday. We note that n(e) + n(e ) = n(s), and dividing each term by n(s), we see that This formula is true in general. In our example, p(e) + p(e ) = 1. p(e ) 1.992 =.008
Thirty people are chosen at random. What is the probability that they all have different birthdays? What is the probability that at least two of the thirty people have the same birthday? Assume a year has 365 days (so we are ignoring the complication of leap years.) n(s)=the number of ways thirty people can have birthdays = 365 30. Let n(e)=the number of ways thirty people can have different birthdays. n(e) = 365 P 30. So we have n(s) = 365 P 30.294 (365) 30 If you use a TI calculator, you would push 365, MATH, PRB, then npr, then 30, ENTER, and then divide by 365 30 365!. When you calculate the numerator 365 P 30 = (365 30)! = 365!, if you attempt to calculate 333! 365! and then divide by 333!, you will get overflow. If you are working near a computer that is running and don t have a calculator handy, the following BASIC code, inserted into a procedure as we have done before (run Excel, click on tools, macro, Visual Basic Editor, insert module, then insert procedure, name the procedure, then insert the code, then run the procedure), will compute n(e): a=1 For i=0 to 29 a=a*(365-i)/365 Next i Cells(1,1)=a Let E be the event that at least two people have the same birthday. Since p(e) + p(e ) = 1, p(e ) 1.294 =.706 This means that if you run the computer simulation of the experiment 1000 times (see the Friday assignment,) you expect that approximately 706 times at least two people will have the same birthday.
We already mentioned that p(e) + p(e ) = 1. More basic rules: E F denotes the event that an event from E or from F (or both) occurs. E F denote the event that an event from E and F occurs. If E and F are mutually exclusive, If E and F are not mutually exclusive, p(e F ) = p(e) + p(f ). p(e F ) = p(e) + p(f ) p(e F ). Example: 1) One card is drawn from a deck. a) What is the probability that it is either a king or a queen? n(e)=8 since there are four kings and four queens. n(s)=52, so /n(s) = 8/52 = 2/13 b) What is the probability that it is neither a king or a queen? If we simply use the above observation, p(e) = 1 p(e ) = 1 2/13 = 11/13 2) Suppose you draw one card from a deck. Let E be the event that the card is red, and F be the event that the card is black. E F represents the event that the card is red or black. E and F are mutually exclusive, and
1 = p(e F ) = p(e) + p(f ) = 1/2 + 1/2. 3) Now let E be the event that the card is red, and F be the event that the card is a king or queen. E and F are not mutually exclusive, and p(e F ) is the probability that a card is a red king or queen. n(s)=52 = the number of ways of choosing a card. There are 4 cards that are red and kings or queens, so p(e F ) = 4/52. Also, p(f ) = 8/52 since there are 8 cards in a deck that are kings or queens. E F represents the event that the card is red or a king or queen. The word or is not exclusive. That means if a card is red and a king, for example, then that is an event in E F. p(e F ) = p(e) + p(f ) p(e F ) = 1/2 + 8/52 4/52 = 15/26. In poker (you are dealt five cards), what is the probability of being dealt a royal flush? (an ace, king, queen, jack, and ten all of the same suit). The number of ways of being dealt five cards is 52 C 5. There are 4 ways you can get a royal flush (once for each suit). p(e) = 4/( 52 C 5 ) = 4/2598960 = 1/649740 4 shoes are selected at random from a pile containing 8 left shoes and 12 right shoes. What is the probability of ending up with two pairs of shoes? A pair of shoes will mean any left shoe and any right shoe. You must choose 2 shoes from the 8 left shoes and 2 shoes from the 12 right shoes. The number of ways of doing this is n(e) = 8 C 2 12 C 2 = 1848 The total number of ways of choosing 4 shoes from the pile of 20 shoes is 20 C 4 = 4845
Then the probability /n(s) = 1848/4845 = 616/1615 In a five card hand, what is the probability of getting exactly 4 cards of the same suit? n(e) is the number of ways of choosing a suit (4) and then choosing exactly 4 cards from that suit, and then choosing the last card from the rest of the deck. Having chosen a suit, there are 13 C 4 ways of choosing four cards from that suit. The remaining card must be chosen from the 39 other cards (of a different suit). Therefore n(e) = 4 13 C 4 39. The total number of ways of choosing 5 cards from the deck is 52 C 5. n(s) = 4 13C 4 39 52C 5 = 143 3332 =.042917... In a five card hand, what is the probability of getting exactly 4 of a kind, e.g, four aces,..., or four kings? There are 13 ways of choosing which kind: ace, two, three,...,king. Having made that choice, there are 48 ways to choose the fifth card. So n(e) = 13 48. The total number of ways of choosing 5 cards from the deck is 52 C 5. n(s) = 13 48 52C 5 = 1 4165 End of Lecture