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probability function we have assuming to be
countably additive ah but countably additive
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ah axiom implies that if we have a disjoint
sequence then the probability of union is
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equal to the sum of the probabilities what
if if we dont have disjoint sequence for example
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if i have two sets say a and b then we have
probability of a union b is equal to probability
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a plus probability b minus probability of
a intersection b so that means if i remove
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probability of a intersection b from there
then we get probability of a union b less
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than or equal to probability a plus probability
of b this is called subadditivity so if in
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general if we consider any sequence of sets
then the probability of union will be less
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than or equal to the sum of the probabilities
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so we have subadditivity of the probability
function now we can stated in the form of
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a theorem for a one a two a n belonging to
b probability of union a i i is equal to one
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to n is less than or equal to sigma probability
of a i i is equal to one to n so one can prove
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this by induction because for n is equal to
one the result is true and if we look at for
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n is equal to two it is already shown to be
true so for n is equal to one the inequality
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is trivially true for n is equal to two which
we will require for extension from k to k
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plus one case so for n is equal to two the
inequality follows
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from the addition rule
so assume it to be true for say n is equal
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to k
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now for n is equal to k plus one we can write
probability of union a i i is equal to one
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to k plus one as less than or equal to probability
of union a i i is equal to one to k plus probability
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of a k plus one by using the result for n
is equal to two so now on this we can make
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the use of assumption that up to n is equal
to k it is true so it becomes less than or
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equal to probability of a i i is equal to
one to k plus probability of a k plus one
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which is nothing but the sum of the probabilities
i is equal to one to k plus one therefore
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by induction the result is true for all of
them
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now if we want to prove the result for a countable
number of these then we can consider the decomposition
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so if we have for any countable sequence say
a i in b probability of union a i i is equal
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to one to infinity is less than or equal to
sigma probability of a i i is equal to one
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to infinity in order to prove this one one
may consider the decomposition of union a
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i into a disjoint decomposition in the following
way let us define say b one is equal to a
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one b two is equal to a two minus a one b
three is equal to a three minus a one union
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a two and so on in general b n is equal to
a n minus union of a i i is equal to one to
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n minus one
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if we consider a venn diagram then it will
be clear that what sets we are defining suppose
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these sets are a one a two a three a four
etcetera then a one and then a two minus a
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one is this set then a three minus a one union
a two becomes this set a four minus a one
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union a two union a three becomes this set
so naturally you can see here that we are
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considering the union as a disjoint union
so then b n is a disjoint sequence of sets
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further union of a i i is equal to one to
infinity is equal to union of b i i is equal
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to one to infinity to prove this let us observe
that union of b i is already a subset of union
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b i because each of the b is is a subset of
the corresponding a is now any point of a
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i let us consider say x belonging to union
of a i let j be the smallest index so that
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x belongs to a j then x will belong to b j
consequently x will belong to union of b is
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as a result since ah union of b i is already
a subset of union ah of a i we are now getting
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union of a i is a subset of union of b i therefore
we must have union of a i is equal to union
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of b is so now if we consider probability
of union of a i i is equal to one to infinity
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it is probability of union of b i i is equal
to one to infinity which is less than or equal
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to which is actually equal to sum of the probability
of b is because b is are now disjoint and
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we can use the axiom of countable additivity
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now each b i is a subset of a i therefore
probability of each b i is less than or equal
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to probability of a i therefore this becomes
less than or equal to sigma probability of
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a i i is equal to one to infinity this proves
the countably additivity of the probability
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function
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ah we also have something called bonferroni
inequalities which basically give the that
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the probability of the unions are bounded
between two bounds
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so for any events a one a two a n in b probability
of the union which is already less than or
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equal to sum of the probabilities it is however
greater than or equal to probability of minus
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ah let me not prove it here the proof will
be by induction
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ah we can see the right hand side has already
been proved to prove the left hand side if
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we take n is equal to one then it is trivially
true for n is equal to two there is equality
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by the addition rule so assuming for n is
equal to k if we write for n is equal to k
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plus one then we can split it into two terms
that is union of a i i is equal to one to
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k union a k plus one on that we apply the
addition rule and then apply the assumption
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for k that will prove the general bonferroni
inequality
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in a similar way we have what is known as
booles inequality the booles inequality gives
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a relation between the intersection likewise
for example if i have a n is any sequence
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of sets in b then probability of intersection
a i i is equal to one to infinity is greater
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than or equal to one minus sigma probability
of a i complement i is equal to one to infinity
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to prove this we simply use the subadditivity
because we can write probability of intersection
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a i i is equal to one to infinity as one minus
probability of intersection a i complement
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now this can be written as one minus probability
of union a i complement by using de morgans
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laws at this is stage we can use the countable
subadditivity so this will become greater
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than or equal to one minus sigma probability
of a i complement [vocalized-noise] ah let
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me give some examples of applications of basic
rules of probability
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let me start from a birthday problem
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suppose there are n persons in a party assuming
that the number of persons is less than or
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equal to three sixty five
and no person has birthday on twenty ninth
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february what is the probability that at least
two persons share the same birthday
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now in order to analyze this problem let us
consider the set theoretic description let
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us consider a to be the event that at least
two
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persons share the same birthday
then if you look at this event it is slightly
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complicated event in the sense that ah two
persons may share three persons may share
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and so on and finding out the probabilities
of each them may be a little bit complicated
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because if we say two persons share then which
of the dates and all others must be on some
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other dates and they should not be the same
suppose we say three persons share then which
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of the three one of the three sixty five days
and all other persons must be on distinct
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days which distinct days so this is a complicated
way to analyze
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however if we use the ah set theoretic representations
we can look at the complementary event a complement
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this means no two persons have the same birthday
now this becomes somewhat simpler because
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if we look at the probability of a complement
assuming all the birthdays to be equally likely
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this number will be simply three sixty five
p n divided by three sixty five to the power
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n here the denominator denotes the total number
of possibilities for n persons to have birthdays
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because each person can have any of the three
sixty five days as a possibly birthday and
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therefore n persons can have possible number
of birthdays as three sixty five to the power
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n
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if we make the assumption that none of them
have the same birthday then it becomes a problem
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of choosing n numbers out of three sixty five
which are distinct so it is nothing but the
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number of permutations taken n at a time from
three sixty five that is equal to three sixty
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five into three sixty four up to three sixty
five minus n plus one divided by three sixty
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five to the power n which we may write as
as a way of representation as one one minus
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one by three sixty five one minus two by three
sixty five and so on up to one minus n minus
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one by three sixty five and so probability
of a becomes one minus
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the product given by these terms ah an interesting
thing would be to look at that how many people
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are required so that at least two will share
a birthday ah if we think from a layman point
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of view then we may think that the numbers
since the number of possible birthdays is
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three sixty five to the power n so n should
be somewhat large in order that this probability
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is significant
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so let us look at the table of probabilities
let us consider say
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n probability of a complement and probability
of a so simple calculation table can be prepared
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if i have n is equal to ten then the probability
of a complement is point eight seven one and
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consequently probability of a becomes point
one to nine if we take n is equal to twenty
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probability of a complement is point five
eight nine and probability of a becomes point
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four one one
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if we take n to be twenty three then probability
of a complement is point four nine three and
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probability of a becomes point five zero seven
that means with as less as only twenty three
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persons the probability that at least two
share a common birthday is more than fifty
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percent so it is from a laymans thinking this
is counter intuitive we need very few persons
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to at least two of them to share a common
birthday
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if i take n is equal to thirty then this probability
becomes point seven zero six if we take fifty
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the probability is point nine seven and for
n is equal to sixty the probability is point
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nine nine four its nearly one that means in
a set of sixty people the probability is nearly
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one that at least two of them will share a
common birthday
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so here you can see that the ah elementary
rules of probability have been used for calculation
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for example we have used the property of the
complementation to evaluate the actual probability
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we have used the method of classical probability
by assuming all the birthdates to be equally
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likely for all the persons [vocalized-noise]
let us look at some other applications of
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the basic rules of the probability
suppose a die is tossed three times independently
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and the outcomes are recorded as numbers a
b and c what is the probability that the roots
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of equation a x square plus b x plus c is
equal to zero are real
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now if we want to calculate this probability
here the outcomes a b and c are random each
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of the values of a b and c can be numbers
one two up to six therefore the quadratic
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equation a x square plus b x plus c is equal
to zero will have the real roots if b square
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minus four a c is positive so we have to look
at the number of cases where b square minus
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four a c is greater than or equal to zero
so this has to be done through an enumeration
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and we can prepare the table that what are
the possibilities of b and therefore the corresponding
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values of b square what are the possible values
of a and c which lead to four a c being less
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than or equal to b square so let us take say
b is equal to one then b square is equal to
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one that means there is no case which will
give me four a c to be less than or equal
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to b square so there is no possibility here
so if you look at the number of cases this
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is zero
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if we take b is equal to two then b square
is equal to four and if we i consider a and
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c to be one one then four a c will become
four so there is one case which will give
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me b square greater than or equal to four
a c if we consider b is equal to three then
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b square is equal to nine now one one one
two and two one there are three cases which
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will give me b square greater than or equal
to four a c if we have b is equal to four
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then b square is equal to sixteen we will
have the cases one one one two two one two
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two which will correspond to four so one four
four one one three three one one two three
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four five six seven eight cases are there
which will give me b square greater than or
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equal to four a c
if we have b is equal to five then b square
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is equal to twenty five then all the above
cases that is eight cases plus we will also
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have one five five one and possibly ah one
four four one so we will will also have one
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six six one two three three two basically
one two three four five six more cases so
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fourteen cases are there if i have b is equal
to six then b square becomes thirty six and
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all the fourteen cases plus we will also have
two four four two then two five is not possible
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three three i think three three must have
come here itself because no it will not come
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here three three will come here because this
will give me nine so there are seventeen cases
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so if you look at the total number of cases
it is thirty nine forty two forty three cases
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are there total number of cases is forty three
and the total number of possibilities if i
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define a to be the event that the roots are
real then the probability of that will be
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given by the favorable number of cases divided
by the total number of cases which is six
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cube here because three dies each of them
have six possibilities so the total number
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of possibilities are six cube there is forty
three by two one six likewise in this problem
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we may also find out the probability of the
ah quadratic equation to have complex roots
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or the real roots but equal etcetera we may
consider all types of possibilities so i will
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end todays lecture by this
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thank you