lecture1 and including sets,probalistic models Flashcards
subset , equall set
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
universal set or omega
contains all those element which are in intrest of taking an experient
complemetn of universal set is empty set or not
yes
partition of set and also define the disjoint
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
set algeba rules
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.
probablity law
a law which assign a non negative number P(A) to our event A in a sample space omega
a sample space is mutually exclusive and collectively exhaustive what does it means
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
sequatial model and how to solve
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
probablity axioms
1) P(A) ≥ 0
2) if a and b are disjoint
P(A ∪ B )=P(A)+P(B ).
3) P(Ω) = 1.
prove that the prob of 0 is zero
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.
Discrete Probability Law
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 }.
equally likely means
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
Discrete Uniform Probability Law
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
the prob of singel event is ——
0
prob properties
(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 )
conditional prob law`
conditional probability of A given B , denoted by P(A | B ).
probablity of intersection is ? here we have two cases
1) if two sets are dependent
2) if two sets are independent
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
in conditional prob the P(Ω | B ) is equall to
P(Ω ∩ B )/p(B) =P(B )/P(B ) = 1
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————-
P(A ∩ B )=P(B )P(A | B )
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
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
what is multiplication rule according to the prof notes
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 .
what is permutation , formual and where we can use it
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.
subset,with ordering,without ordering
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
possible ways to partition n element in k subset of cardnility nk
n!/n1!………nk!
example : Imagine to have a box containing n balls. Then there are:
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
subset with repetition
two parts 1)with ordering,without ordering
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)!
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:
1)n^k order ways to pick k
2)(n+k−1)!/k!(n+k−1)! unorder way to pick k
