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

2. Mathematical Expectation مثا ل قيمة Y 13 المجموع P(y)3/41/41 Y p(y)3/4 6/4

3. Variance of a random Variable V(X) = E(X 2 ) – [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 } 23 0 2 2 1 1 1 3210X 9410X2X2 1/83/8 1/8F(X)

4. SOLUTION: V(X) = E(X 2 ) – [E(X)] 2 V(X) = E(X 2 ) – [E(X)] 2 3 E(X 2 )=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 مثا ل654321 (1,6)(1,5)(1.4)(1,3)(1,2)(1,1)1 (2,6)(2,5)(2,4)(2,3)(2,2)(2,1)2 (3,6)(3,5)(3,4)(3,3)(3,2)(3,1)3 (4,6)(4,5)(4,4)(4,3)(4,2)(4,1)4 (5,6)(5,5)(5,4)(5,3)(5,2)(5,1)5 (6,6)(6,5)(6,4)(6,3)(6,2)(6,1)6 Elements of the sample space = 6 2 = 36 elements X is a random variable defined in S The range of it is {2,3,4,……….,11,12}

6. قيمة X 23456789101112 P(X =x) 1/362/363/364/365/366/365/364/363/362/361/36 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 مثا ل قيمة X 0.270 0.331 0.162 0.143 0.104 1 المجموع 0 0.33 0.32 0.42 0.40 1.47

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

10. Course 3 Factorial 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!54321 5! = 5 4 3 2 1 Read 5! as “five factorial.” Reading Math

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

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

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

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

15. Course 3 10-9 Permutations and Combinations Permutation Q: What is a permutation? A permutation is an arrangement of things in a certain order. first letter ? second letter ? third letter ? 3 choices2 choices1 choice The product can be written as a factorial. 3 2 1 = 3! = 6

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

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

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

19. Course 3 10-9 Permutations and Combinations

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

21. 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. 12 possible DVDs 4 DVDs chosen at a time 12! 4!(12 – 4)! = 12! 4!8! = 12 11 10 9 8 7 6 5 4 3 2 1 (4 3 2 1)(8 7 6 5 4 3 2 1) 12 C 4 = = 495

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

23. Jan 2009 23

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 1.The expected value is equal to: 2.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) = c n x p x q n-x Given that,x=2 p = 0.8, n=4 = 1-p q =1-0.8 = 0.2 f(2)= c 4 2 p 2 q 4-2 2 0.8) 2 (0.2))4! 2!2!

29. . 4.3.2!(0.64)(0.4) = 0.1536 2!2! 2. Achieving the goal at least twice المطلوب: p(x>=2) = f(2) + f(3)+ f(4) 0.9728 = =0.1536 + c 4 3 (0.8) 3 (0.2)+ c 4 4 (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) = c n x p x q n-x X = 0,1,2 ………20 calculate P(x>=2) or 1-p(x<2) = 1 – [P(0)+ P(1) ] = 1- [f(0) + f(1)] 1-[c 20 0 (0.1) 0 (0.9 20 )] + [c 20 1 (0.1) 1 (0.9) 9 ]