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
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.
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
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]