Lecture No. 4

•Combined Sample space

•Permutation

•Combinations

•Bernoulli Trials

Combined Sample Space

First, we’ll do some permutation problems.

Permutations are “arrangements”.

Let’s do a permutation problem.

How many different arrangements are there for 3 books on a shelf?

Books A,B, and C can be arranged in these ways:

ABC ACB BAC BCA CAB CBA

Six arrangements or 3! = 3x2x1 = 6

In a permutation, the order of the books is important.

Each different permutation is a different arrangement.

The arrangement ABC is different from the arrangement CBA, even though they are the same 3 books.

You try this one:

1. How many ways can 4 books be arranged on a shelf?4! or 4x3x2x1 or 24 arrangements

Here are the 24 different arrangements:

ABCD ABDC ACBD ACDB ADBC ADCB

BACD BADC BCAD BCDA BDAC BDCA

CABD CADB CBAD CBDA CDAB CDBA

DABC DACB DBAC DBCA DCAB DCBA

Now we’re going to do 3 books on a shelf again, but

this time we’re going to choose them from a group of

8 books.We’re going to have a lot more

possibilities this time, because there are many groups of 3 books to be chosen from the total 8, and there are several different arrangements

for each group of 3.

If we were looking for different arrangements for all 8 books,

then we would do 8!

But we only want the different arrangements for

groups of 3 out of 8, so we’ll do a partial factorial,

8x7x6

=336

Try these:

1. Five books are chosen from a group of ten, and put on a bookshelf. How many possible arrangements are there?

10x9x8x7x6 or 30240

2. Choose 4 books from a group of 7 and arrange them on a shelf. How many different arrangements are there?

7x6x5x4 or 840

Permutations

To find the number of Permutations of n items chosen r at a time, you can use the formula

. 0 where nrrn

nrpn

)!(

!

603*4*5)!35(

!535

2!

5! p

Now, we’ll do some combination problems.

Combinations are “selections”.

There are some problems where the order of the items is NOT important.

These are called

combinations.You are just making selections, not making different arrangements.

Example: A committee of 3 students must be selected from a group of 5 people. How many possible different committees could be formed?

Let’s call the 5 people A,B,C,D,and E.

Suppose the selected committee consists of students E, C, and A. If you re-arrange the names to C, A, and E, it’s still the same group of people. This is why the order is not important.

Because we’re not going to use all the possible combinations of ECA, like EAC, CAE, CEA, ACE, and AEC, there will be a lot fewer committees.

Therefore instead of using only 5x4x3, to get the fewer committees, we must divide.

5x4x3

3x2x1

(Always divide by the factorial of the number of digits on top of the fraction.)

Answer:

10 committees

Now, you try.

1. How many possible committees of 2 people can be selected from a group of 8?

8x7

2x1or 28 possible committees

2. How many committees of 4 students could be formed from a group of 12 people?

12x11x10x9

4x3x2x1

or 495 possible committees

Combinations

To find the number of Combinations of n items chosen r at a time, you can use the formula

. 0 where nrrnr

nrCn

)!(!

!

21

Bernoulli Trials

•Suppose an experiment with two possible outcomes, such as tossing a coin.

•Each performance of such an experiment is called a Bernoulli trial.

•We will call the two possible outcomes a success or a failure, respectively.

•If p is the probability of a success and q is the probability of a failure, it is obvious thatp + q = 1.

•The trials are independent.

22

Bernoulli Trials•Theorem: The probability of k successes in n independent Bernoulli trials, with probability of success p and probability of failure q = 1 – p, is•C(n, k)pkqn-k .

•We denote by b(k; n, p) the probability of k successes in n independent Bernoulli trials with probability of success p and probability of failure q = 1 – p.•Considered as function of k, we call b the binomial distribution.

.

When N or (N-K) is Large• When N, K or (N-K) is large it is difficult to

calculate the factorial. In this case approximation are useful. One of which is Stirling’s formula which is stated below.

• This is exact when ‘n’ approaches infinity.

nn

n! 2πne

~

De-Moivre Laplace Approximation

This formula holds for large values of N,K and (N-K). K near np such that its deviation from np are small in magnitude relative to both np and nq.This approximation fails when n is large and p is small. In this case we may use poisson aproximation.

Attention

• Chapter 1 is over.