100% satisfaction guarantee Immediately available after payment Both online and in PDF No strings attached
logo-home
Previously searched by you
Previously searched by you
logo
A course in probability theory and and statistics for beginners $5.50
Add to cart
Quickly navigate to
Preview
  2.  
  3. Stellenbosch University (SUN)
  4.  
  5. Probability Theory And Stats - 144 (PS144)

Class notes

A course in probability theory and and statistics for beginners
 80 views  2 purchases
  • Course
  • Probability Theory And Stats - 144 (PS144)
  • Institution
  • Stellenbosch University (SUN)
  • Book
  • A First Course in Probability

This material covers important introductory concepts in statistics, such as counting principles, permutations and combinations, axioms of probability, conditional probability, probability distributions and a few well-known discrete and continuous probability distributions (e.g. binomial distributio...

[Show more]

Preview 4 out of 43  pages

Document information

  • June 18, 2023
  • 43
  • 2022/2023
  • Class notes
  • Dr. trudy sandrock
  • All classes

Subjects

  • probability theory and statistics

Connected book

book image

Book Title:

Author(s):

  • Edition:
  • ISBN:
  • Edition:

More summaries for

  • Solutions Manual A First Course in PROBABILITY Seventh Edition Sheldon Ross
  • Summary Probability Theory and Stats 114
  • A First Course in Probability Solution Manual
All for this textbook (4)

Written for

  • Stellenbosch University (SUN)
  • Probability Theory And Stats - 144 (PS144)
All documents for this subject (1)

Seller

avatar-seller
tanweerkola

Content preview

COMBINATORIAL ANALYSIS



}
↳
Theory that allows us to count the This is useful for calculating probabilities

number of outcomes of an
experiment .




The basic principle of counting
If experiment 1 has m possible outcomes and for each of these outcomes , experiment 2


has n possible outcomes ,
then the total number of outcomes for the whole experiment is mn .




experiment 1
①


y
outcome 1
12 outcomes )
)
outcome 1

"" "
" " °" " ② " "




§
possible outcomes
outcome
③
1



outcome 2



outcome 2
④
experiment 2
( 2 outcomes for each outcome


in experiment 1)



The tree diagram illustrates how we can find the total number of outcomes .
However ,

this could also be calculated the basic outcomes
using counting principle 2×2 4
=
as .




↳
This the theorem
is a
very trivial example but it illustrates




Suppose have each of whom have children If need to choose
Example we 7- women
,
3 .
we


one woman and one of her children for a free holiday ,
how many possible choices

are there ?




Experiment 1 → Choice of the mother ( 7- outcomes )


Experiment 2 → Choice of the child for the chosen mother ( 3 outcomes )


i. Total outcomes = 7×3 =
21 outcomes

,The generalised basic principle of counting

suppose we need to perform r experiments ,
where the first experiment has n
,
outcomes ,

the second experiment has nz Outcomes ,
. . .



,
the rth ( last ) experiment has nr outcomes .




The total number of outcomes of the r experiments is n
,
✗
nzx .
. . ✗ nr .




'

- -




Example A student body consists of 4 freshmen , 3 sophomores , 2
juniors and 5 seniors .




If a student council is to be formed and must consist of 1 person from each

class 14 people on the council in total ) ,
how many different committees can



be formed ? " "
A useful
"
trick
"
to when have AND and
is multiply we




✗
"
to add when we have OR "
.




We require : 1 freshman AND i sophomore AND I
junior AND I senior


From the generalised of possible committees
counting
basic principle : 4×3×2×5 = 120




Example How many 7 -



digit license plates can we make if the first 3
digits are to be letters and


the remaining 4 digits by numbers ? What if repetition of letters and numbers is not allowed ?




Ii ) We have : 26×26 ✗ 26 ✗ 10×10×10×10


= 263 ✗ 104

=
175 760 000 possible license plates




Iii ) In this still the generalised principle of have
case ,
using counting ,
we



26 ✗ 25×24 ✗ 10×9×8×7 =
78 624 000 possible license plates
↑



}
we can't use the one letter
think of how options each
we chose in position 1 ,
many
and so on .

position has available .

, PERMUTATIONS

↳ Deals with ordered
arrangements of objects



First result

If want to arrange objects where each object in every then the number
we n , appears arrangement ,




the number of arrangements we can make is n ! .




'



-




Example How
many different batting orders are possible for a 9-
player baseball team ?




There are 9 ! =
362880 possible batting orders




Example If a student has 4 maths books , 2 chemistry books and 3 biology books ,
how
many

these books bookshelf if the books to the
ways can we
arrange on a
belonging same



subject should each distinct ?
be
grouped together and book is




There are 4 ! arrangements of math books , 2 ! arrangements of Chem books and 3 !
arrangements of

bio books . We then regard each group as an object giving ,
us 3 !
ways to arrange each of the groups .




i. (4! ✗ 2 ! ✗ 3 ! ) ✗ 3 ! =
1728 arrangements
the
I ↑
arranging
each
books within each arranging group
particular subject




second result

If we have n objects with n
, objects identical to each other , nz objects identical to each other ,




nr Objects identical to each other then the number of these objects can be
arranged
ways
- .

,
_



,




is calculated as :




can think of it as
n !
"
"

removing the repetitions
n, ! nz ! .
. .

nr ! I

, Example If we have 5 red balls ,
3 white balls and 6 blue balls ,
how many different ways can


these balls in line ?
we
arrange a




There (5+3+6) ! !
are =
14
= 168168 possible distinct arrangements
5 ! 3 ! 6 ! 5 ! 3 ! 6!




Further result

The number of permutations of r different objects which are chosen from n objects is


defined as :




npr =
Prn = n In 1) ( n 2)
- -


. . . (n -
r +
1) =
n !


( n -
r )!




L 0 Think of it as we want to select r objects from
a
larger group of n objects and the objects are ordered .




How there to arrange students in from group of
a row
Example many ways are 5 a



10 students ?




10ps = 10 ! =
10 ! =
10×9 ✗ 8×7×6×5 ✗ 4×3×2×1
= 10×9×8×7 ✗ 6 =
30240


(10-5) ! 5
! 5×4×3×2 ✗ 1




COMBINATIONS

↳ Similar to but when not important
permutations , ordering is .




Combination definition

we define the combination where we want to choose r objects from the larger
of objects and the ordering of the objects is not important
group n as :




(f) for
=
n ! r ≤ n

(n -
r )!r!

The benefits of buying summaries with Stuvia:

Guaranteed quality through customer reviews

Guaranteed quality through customer reviews

Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.

Quick and easy check-out

Quick and easy check-out

You can quickly pay through credit card or Stuvia-credit for the summaries. There is no membership needed.

Focus on what matters

Focus on what matters

Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!

Frequently asked questions

What do I get when I buy this document?

You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.

Satisfaction guarantee: how does it work?

Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.

Who am I buying these notes from?

Stuvia is a marketplace, so you are not buying this document from us, but from seller tanweerkola. Stuvia facilitates payment to the seller.

Will I be stuck with a subscription?

No, you only buy these notes for $5.50. You're not tied to anything after your purchase.

Can Stuvia be trusted?

4.6 stars on Google & Trustpilot (+1000 reviews)

50843 documents were sold in the last 30 days

Founded in 2010, the go-to place to buy study notes for 14 years now

Start selling
$5.50  2x  sold
  • (0)
Add to cart
Added