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  5. Probability Theory And Stats - 144 (PS144)

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A course in probability theory and and statistics for beginners
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  • 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...

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Document information

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

Subjects

  • probability theory and statistics

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  • Probability Theory And Stats - 144 (PS144)
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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!

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