1. 1 Section 5.1 Discrete Probability

2. 2 LaPlace’s definition of probability Number of successful outcomes divided by the number of possible outcomes This definition works when all outcomes are equally likely, and are finite in number

3. 3 Finite Probability Experiment: a procedure that yields one of a given set of possible outcomes Sample space: set of possible outcomes Event: a subset of the sample space LaPlace’s definition, stated formally, is: The probability p of an event E, which is a subset of a finite sample space S of equally likely outcomes is p(E) = |E|/|S|

4. 4 Example 1 What is the probability that you will draw an ace at random from a shuffled deck of cards? There are 4 aces, so |E| = 4 There are 52 cards, so |S| = 52 So p(E) = |4|/|52|, or 1/13

5. 5 Example 2 What is the probability that at randomly- selected integer chosen from the first hundred positive integers is odd? S = 100, E = 50 So p(E) = 1/2

6. 6 Example 3 What is the probability of winning the grand prize in the lottery, if to win you must pick 6 correct numbers, each of which is between 1 and 40? There is one winning combination The total number of ways to choose 6 numbers out of 40 is C(40,6) = 40!/(34!6!) So your chances of winning are 1/3,838,380

7. 7 Example 4 What is the probability that a 5-card poker hand does not contain the ace of hearts? If all hands are equally likely, the probability of a hand NOT containing a particular card is the quotient of: –probability of picking 5 cards from the 51 remaining: C(51,5) and –probability of picking any 5 cards from entire deck: C(52,5)

8. 8 Example 4 C(51,5) = 51!/5!46! C(52,5) = 52!/5!47! So C(51,5)/C(52,5) = (51!/5!46!)(5!47!/52!) Through cancellation, we get: 47/52 or ~.9

9. 9 Example 5 There are C(52,5) = 52!/(47!5!) = 2,598,960 possible hands of 5 cards in a deck of 52 What is the probability of getting 4 of a kind in a hand of 5?

10. 10 Example 5 Using the product rule, the number of ways to get 4 of a kind in a hand of 5 is the product of: –the number of ways to pick one kind: C(13,1) –the number of ways to pick 4 of this kind from the total number in the deck of this kind: C(4,4) –the number of ways to pick the 5th card: C(48,1) So the probability of being dealt 4 of a kind is: (C(13,1)C(4,4)C(48,1))/C(52,5) = 13*1*48/2,598,960 or.00024

11. 11 Example 6 What is the probability of a 5-card poker hand containing a full house (3 of one kind, 2 of another)? By the product rule, the number of hands containing a full house is the product of: –ways to pick 2 kinds in order: P(13,2): order matters because 3 aces, 2 tens  3 tens, 2 aces –ways to pick 3 out of 4 of first kind: C(4,3) –ways to pick 2 out of 4 of second kind: C(4,2 )

12. 12 Example 6 P(13,2) * C(4,3) * C(4,2): –P(n,r) = n(n-1) * … * (n-r+1); since 13-2 = 11, P(n,r) = 13 * 12 –C(4,3) = 4!/3!1! & C(4,2) = 4!/2!2! = 24/6 & 24/4 –So result is 156 * 4 * 6 = 3744 Because there are 2,598,960 possible poker hands, the probability of a full house is 3744/2598960 = ~.0014

13. 13 Example 7 What is the probability that a 5-card poker hand contains a straight (5 consecutive cards, any suit)? Assuming ace is always high, the 5 cards could start with any of: {2,3,4,5,6,7,8,9,10} So there are C(9,1) or 9 ways to start There are 4 suits, so there are 4 cards of each kind, so there are C(4,1) or 4 ways to make each of the 5 choices

14. 14 Example 7 Putting this information together with the product rule, there are 9 * 4 5 = 9,216 different possible hands containing a straight Since there are 2,598,960 hands possible, the probability of a hand containing a straight is: 9,216/2,598,960 = ~.0035

15. 15 Complement of an event Let E be an event in sample space S. The probability of E, the complementary event of E is: p(E) = 1 - p(E) Note that |E| = |S| - |E| So p(E) = (|S| - |E|)/|S| = 1-|E|/|S| = 1- p(E) Sometimes it’s easier to find the probability of a complement than the probability of the event itself

16. 16 Example 8 A sequence of 10 bits is randomly generated. What is the probability that at least one bit is 0? –E: at least 1 of 10 bits is 0 –E: all bits are 1s –S: all strings of 10 bits

17. 17 Example 8 Since p(E) = 1-p(E) and there are 2 10 possible bit combinations, but only 1 containing all 1s: p(E) = 1/2 10 = 1-1/1024 = 1023/1024, or 99.9% probability that at least one bit is 0 in a random 10-bit string

18. 18 Probability of Union of 2 Events The probability of the union of 2 events is the sum of the probabilities of each of the events, less the probability of the intersection of the 2 events: p(E 1  E 2 ) = p(E 1 ) + p(E 2 ) - p(E 1  E 2 )

19. 19 Probability of Union of 2 Events: proof of theorem Recall the formula for the number of elements in the union of 2 sets: |E 1  E 2 | = |E 1 | + |E 2 | - |E 1  E 2 | So, p |E 1  E 2 | = |E 1  E 2 |/|S| =(|E 1 | + |E 2 | - |E 1  E 2 |)/|S| = |E 1 |/|S| + |E 2 |/|S| - |E 1  E 2 |/|S| = p(E 1 )+p(E 2 )-p(E 1  E 2 )

20. 20 Example 9 What is the probability that a positive integer selected at random from the set of positive integers not exceeding 100 is divisible by either 2 or 5? E 1 = selected integer divisible by 2; |E 1 |=50 E 2 = selected integer divisible by 5; |E 2 |=20

21. 21 Example 9 So E 1  E 2 is the event that the number is divisible by either 2 or 5 and E 1  E 2 is the event that the number is divisible by both (divisible by 10): |E 1  E 2 |=10 So p(E 1  E 2 ) = p(E 1 ) + p(E 2 ) - p(E 1  E 2 ) = 50/100 + 20/100 - 10/100 = 60/100 = 3/5

22. 22 Probabilistic reasoning … is determining which of 2 events is more likely The Monty Hall 3-door puzzle is an example of such reasoning: –select one of 3 doors, one of which has the GRAND PRIZE!!!!! behind it –once selection is made, Monty opens one of the other doors (knowing it is a loser) –then he gives you the option to switch doors -should you?

23. 23 Solution to Monty Hall 3-door puzzle Initial probability of selecting the grand prize door is 1/3 Monty always opens a door the prize is NOT behind The probability you selected incorrectly is 2/3 (since you only had a 1/3 chance of a correct selection)

24. 24 Solution to Monty Hall 3-door puzzle If you selected incorrectly, when Monty selects another door (without prize), the prize must be behind the remaining door You will always win if you chose wrong the first time, then switch So by changing doors, you probability of winning is 2/3 Always change doors!

25. 25 Section 4.4 Discrete Probability - ends -