lecture1 and including sets,probalistic models

Question 1

subset , equall set

Answer 1

If every element of a set S is also an element of a set T ,wesay that S
is a subset of T , and we write S ⊂ T or T ⊃ S .IfS ⊂ T and T ⊂ S , the
two sets are equal, and we write S = T

Question 2

universal set or omega

Answer 2

contains all those element which are in intrest of taking an experient

Question 3

complemetn of universal set is empty set or not

Answer 3

yes

Question 4

partition of set and also define the disjoint

Answer 4

disjoint : if the intersection of two set is zero so disjoint
if we have many sets and they are disjoint but the union of all that set make S so they all are partition of set S

Question 5

set algeba rules

Answer 5

S ∪ T = T ∪ S,
S ∩ (T ∪ U )=(S ∩ T ) ∪ (S ∩ U ),
(S c)c = S,
S ∪ Ω=Ω
S ∪ (T ∪ U )=(S ∪ T ) ∪ U,
S∪ (T ∩ U )=(S ∪ T ) ∩ (S ∪ U ),
S ∩ S c =Ø,
S∩ Ω=S.

Question 6

probablity law

Answer 6

a law which assign a non negative number P(A) to our event A in a sample space omega

Question 7

a sample space is mutually exclusive and collectively exhaustive what does it means

Answer 7

mutually exclusive : the outcome is unique we can not have two outcom from the same sample space
collectvily exhaustive : no matter what happens in our experminet the outcome will be subset of omega that is sample space

Question 8

sequatial model and how to solve

Answer 8

sequantail models are those in which we are repating an exp in a sequace like three time toss a coin or 5 succuve days looking for stock price etc
it can be solved by using tree modle or a table

Question 9

probablity axioms

Answer 9

1) P(A) ≥ 0
2) if a and b are disjoint
P(A ∪ B )=P(A)+P(B ).
3) P(Ω) = 1.

Question 10

prove that the prob of 0 is zero

Answer 10

1=P(Ω) = P(Ω ∪ Ø) = P(Ω) + P(Ø) = 1 + P(Ø),
and this shows that the probability of the empty event is 0:
P(Ø) = 0.

Question 11

Discrete Probability Law

Answer 11

if we have finite number of events in prob model os the prob is equall to the sum of indivduiale prob
P {s1,s2,…,sn } = P{s1} + P{s2} + ··· + P{sn }.

Question 12

equally likely means

Answer 12

if all the element in an event have the same single prob so called equally likely
eg toss a fair coin it means that the total event that is H or T has same prob that is 1/2

Question 13

Discrete Uniform Probability Law

Answer 13

if finite event and equally likely then the prob of event is
P(A)= number of element of A / n
here n = cardinanlity of omega

Question 14

the prob of singel event is ——

Answer 14

0

Question 15

prob properties

Answer 15

(a) If A ⊂ B , then P(A) ≤ P(B ).
(b) P(A ∪ B )=P(A)+P(B ) − P(A ∩ B ).
(c) P(A ∪ B ) ≤ P(A)+P(B ).
(d) P(A ∪ B ∪ C )=P(A)+P(Ac ∩ B )+P(Ac ∩ B c ∩ C )

Question 16

conditional prob law`

Answer 16

conditional probability of A given B , denoted by P(A | B ).

Question 17

probablity of intersection is ? here we have two cases
1) if two sets are dependent
2) if two sets are independent

Answer 17

if the the sets are dependent so we can find the intesection by simply using the formula for conditional prob
p(A and B) = p(A|B) and p(b)
and vise versa when p(b and a ) is requiared
2) if the two events are not depdents so then we have a formal case
where p(A|B) = p(A) and p(b|a) = p(b)
so interstion will be
p(a and b) = p(a).p(b) and vise versa

Question 18

in conditional prob the P(Ω | B ) is equall to

Answer 18

P(Ω ∩ B )/p(B) =P(B )/P(B ) = 1

Question 19

note: when we have an expermnt with sequnetal character so we can find the uncondintional prob by using the conditional prob and the formual that help us is————-

Answer 19

P(A ∩ B )=P(B )P(A | B )

Question 20

no answer just remberb it
if you are given a conditional prob with diffferent stages so remberb this to cal the final prob
Each event of interest corresponds
to a leaf of the tree and its probability is equal to the product of the probabilities
associated with the branches in a path from the root to the corresponding leaf

Answer 20

Each event of interest corresponds
to a leaf of the tree and its probability is equal to the product of the probabilities
associated with the branches in a path from the root to the corresponding leaf

Question 21

what is multiplication rule according to the prof notes

Answer 21

if we have omega1,omega2…..omegaN and we want to take one element from each so how many possible ways we have
so to compute this we intriduce multiplication rule which is the productt of each omega cardianlity
|Ω1 × Ω2 × . . . × ΩN | = n1 n2 · · · nN .

Question 22

what is permutation , formual and where we can use it

Answer 22

there are how many possible wasy of orderning n elemnts to answer this question we will use permutation
as if we have n chioces for 1st element and n-1 choices for second element and n-2 for third so on so overall we have n! choices this is done by permutaion and using multiplication rule already described above.

Question 23

subset,with ordering,without ordering

Answer 23

well here subset means how manay wasy to choose k element from n element
1) with ordering = n(n − 1) · · · (n − k + 1) = n!/(n − k )!
2)without orderding = n!/k!(n − k )!
since we have (n-k)! choices for ordering k and then we have k! for unorder k element to combine them both we fet our desir result

Question 24

possible ways to partition n element in k subset of cardnility nk

Answer 24

n!/n1!………nk!

Question 25

example : Imagine to have a box containing n balls. Then there are:

Answer 25

n!/(n-k)! orders ways to pick k balls
n!/k!(n-k)! ways to pick unorders balls
n!/n1!.n2!.n3! ways to devide into 3 unorder group of n elements

Question 26

subset with repetition
two parts 1)with ordering,without ordering

Answer 26

1) with order and repetition
since we hvae n choices for 1st elem
n choices for second elem and so on so we have n^k choices
2) without ordering but with reputation
(n+k−1)!/k!(n+k−1)!

Question 27

example : Imagine to again have a box with n balls, but now each time a ball is picked, it
is put back in the box (so that it can be picked again). There are:

Answer 27

1)n^k order ways to pick k
2)(n+k−1)!/k!(n+k−1)! unorder way to pick k