20/6/1435 h Sunday

Lecture 11

Jan 2009 1

Mathematical Expectationمثال

المجموع 3 1 YYقيمة قيمة

1 1/4 3/4 P(y)

6/4 3/4 3/4 Y p(y)

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:

الحلS = { HHH , HHT , HTH , THH , TTT , HT , TTH , HTT }

23 022 11 1

X 0 1 2 3X2 0 1 4 9

F(X) 1/8 3/8 3/8 1/8

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

In an experiment of rolling two fair dice, X is defined as the sum of two up facesQ: Construct a probability distribution table

مثال

11 22 33 44 55 66

11 (1,1) (1,2) (1,3) (1.4) (1,5) (1,6)

22 (2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

33 (3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

44 (4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

55 (5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

66 (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

Elements of the sample space = 62 = 36 elements

X is a random variable defined in SThe range of it is {2,3,4,……….,11,12}

1212 1111 1010 99 88 77 66 55 44 33 22 XXقيمة قيمة

1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36 P(X =x)

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

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

مثال

Vocabulary

factorial

permutation

combination

Insert Lesson Title Here

Course 3

10-9Permutations and Combinations

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!5! = 55 • 44 • 33 • 22 • 11

Read 5! as “five factorial.”

Reading Math

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.

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)!

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.

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)!

Course 3

10-9Permutations 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 choices 2 choices 1 choice

The product can be written as a factorial.

• •

3 • 2 • 1 = 3! = 6

Course 3

By definition, 0! = 1.

Remember!

Q: Write a general formula for permutation

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=6P6 =

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.

Course 3

Combinations

A combination is a selection of things in any order.

Q:Define Combination?

Course 3

10-9Permutations and Combinations

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!10C7 =

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.

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)

12C4 =

= 495

مثال

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

Jan 2009 23

Discrete probability distributions

The binomial distribution The Poisson distribution Normal Distribution

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.

Q: Write a general format for the binomial distribution?

.

knk pp )1(

kn

knkknk ppknk

nppk

nkp

)1(

)!(!!

)1()(

Q:Write a general formula for the mathematical expectation and the variance

1. The expected value is equal to:

pkXE )(

2. and variance can be obtained from:

qpkXVar )(

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 p

x q n-x

Given that ,x=2 p = 0.8 , n=4 =1-p q

=1-0.8 = 0.2

f(2)= c42 p

2 q 4-2 2 0.8(2)0.2((4 !

2!2!

.4.3.2!)0.64)(0.4 = (0.1536

2!2!2.Achieving the goal at least twice

p(x>=2) المطلوب: =f(2) + f(3)+ f(4)

0.9728 = =0.1536 + c43 (0.8)3(0.2)+ c4

4 (0.8)4(0.2)0

OR:P(x>=2) = 1- p(x<2)

=1] -f(0) +f(1) [

.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 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-[c200 (0.1)0 (0.920 )] + [c20

1 (0.1)1(0.9)9]