20/6/1435 h Sunday

1. 20/6/1435 h Sunday Lecture 11 Jan 2009

2. Mathematical Expectation مثال

3. Variance of a random Variable V(X) = E(X2) – [E(X)]2 Q: In an experiment of tossing a fair coin three times observing the number of heads x find the varianc of x: الحل S = { HHH , HHT , HTH , THH , TTT , HT , TTH , HTT } 0 1 1 2 1 3 2 2

4. SOLUTION: V(X) = E(X2) – [E(X)]2 V(X) = E(X2) – [E(X)]2 3 E(X2)=0(1/8) + 1(3/8) + 4(3/8) + 9(1/8) = 24/8 = 3 E(X) = ∑ X.F(X) 1.5= ( E(X)=0 (1/8) + 1( 3/8 )+ 2(3/8 )+ 3(1/8 V(X) = 3-(1.5)2 = 3-(2.25) = 0.75

5. In an experiment of rolling two fair dice, X is defined as the sum of two up faces Q: Construct a probability distribution table مثال Elements of the sample space = 62 = 36 elements X is a random variable defined in S The range of it is {2,3,4,……….,11,12}

6. What is the probability that X= 4 i.e what is the probability that the sum of the two upper faces =4

8. Q: Show if the following table is a probability distribution table? If yes calculate the mathematical expectation (mean) of X مثال

9. 10-9 Permutations and Combinations Course 3 Insert Lesson Title Here Vocabulary factorial permutation combination

10. Factorial Reading Math Read 5! as “five factorial.” Course 3 The factorial of a number is the product of all the whole numbers from the number down to 1. The factorial of 0 is defined to be 1. 5! = 5 • 4 • 3 • 2 • 1

11. 8 •7 • 6 • 5 • 4 • 3 • 2 • 1 6 • 5 • 4 • 3 • 2 • 1 Course 3 Example 1 Evaluate each expression. A. 9! 9 •8 •7 • 6 • 5 • 4 • 3 • 2 • 1 = 362,880 8! B. 6! Write out each factorial and simplify. Multiply remaining factors. 8 •7 = 56

12. 10! (9 – 2)! 10! 7! 10 • 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1 7  6  5  4  3  2  1 Course 3 Example 2 C. Subtract within parentheses. 10 • 9 •8 = 720

13. 7 • 6 • 5 • 4 • 3 • 2 • 1 5 • 4 • 3 • 2 • 1 Course 3 Example 3 Evaluate each expression. A. 10! 10 • 9 • 8 •7 • 6 • 5 • 4 • 3 • 2 • 1 = 3,628,800 7! B. 5! Write out each factorial and simplify. Multiply remaining factors. 7 •6 = 42

14. 9! (8 – 2)! 9! 6! 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1 6  5  4  3  2  1 Course 3 Example 4 C. Subtract within parentheses. 9 •8 • 7 = 504

15. 10-9 Permutations and Combinations first letter second letter third letter ? ? ? Course 3 Permutation Q: What is a permutation? A permutation is an arrangement of things in a certain order. 3 choices 2 choices 1 choice • • The product can be written as a factorial. 3 • 2 • 1 = 3! = 6

16. Remember! By definition, 0! = 1. Course 3 Q: Write a general formula for permutation

17. 6! (6 – 6)! The number of books is 6. = = 6! The books are arranged 6 at a time. 0! 6 • 5 • 4 • 3 • 2 • 1 = 1 Course 3 Example 1 Jim has 6 different books. Find the number of orders in which the 6 books can be arranged on a shelf. 6P6 = 720 There are 720 permutations. This means there are 720 orders in which the 6 books can be arranged on the shelf.

18. Combinations Course 3 Q:Define Combination? A combination is a selection of things in any order.

19. 10-9 Permutations and Combinations Course 3

20. 10! 7!(10 – 7)! = 10 possible books 10! = 7!3! 10 • 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1 7 books chosen at a time (7 • 6 • 5 • 4 • 3 • 2 • 1)(3 • 2 • 1) Course 3 Example 1 If a student wants to buy 7 books, find the number of different sets of 7 books she can buy. 10C7 = = 120 There are 120 combinations. This means that Mary can buy 120 different sets of 7 books.

21. 12! 4!(12 – 4)! = 12 possible DVDs 12! 4!8! 12 • 11 • 10 • 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1 = 4 DVDs chosen at a time (4 • 3 • 2 • 1)(8 • 7 • 6 • 5 • 4 • 3 • 2 • 1) Course 3 A student wants to join a DVD club that offers a choice of 12 new DVDs each month. If that student wants to buy 4 DVDs, find the number of different sets he can buy. 12C4 = = 495

22. مثال Q: Calculate the number of ways, we can choose a group of 4 students from a class of 15 students

23. Jan 2009

24. Discrete probability distributions • The binomial distribution • The Poisson distribution • Normal Distribution

25. The binomial distribution • Suppose that n independent experiments, or trials, are performed, where n is a fixed numer, and that each experiment results in „success” with probability p and a „failure” with probability q=1-p. • The total number of successes, X, is a binomial random variable with parameters n and p.

26. Q: Write a general format for the binomial distribution? .

27. Q:Write a general formula for the mathematical expectation and the variance The expected value is equal to: and variance can be obtained from:

28. Examples For a long time it was observed that the probability to achieve your goal is 0.8.If we shoot four bullets at a certain target, find these probabilities:- 1. Achieving a goal twice f(x) = cnx px q n-x Given that ,x=2 p = 0.8 , n=4 = 1-pq =1-0.8 = 0.2 f(2)= c42 p2 q 4-2 20.8)2(0.2))4! 2!2!

29. . 4.3.2!(0.64)(0.4) = 0.1536 2!2! • Achieving the goal at least twice المطلوب: p(x>=2) = f(2) + f(3)+ f(4) 0.9728 = =0.1536 + c43 (0.8)3(0.2)+ c44 (0.8)4(0.2)0 OR: P(x>=2) = 1- p(x<2) = 1- [f(0) +f(1)]

30. . Example 2 In a class containing 20 students, if 90% of students who registered in a statistic course were passed, what is the probability that at least 2 students will fail n = 20, p = 0.1, q= 0.9 f(x) = cnx px q n-xX = 0,1,2 ………20 calculate P(x>=2) or 1-p(x<2) = 1 – [P(0)+ P(1) ] = 1- [f(0) + f(1)] 1-[c200 (0.1)0 (0.920 )] + [c201 (0.1)1(0.9)9]