Problems in Probability - Mathematics, Engineering Video Lecture - Engineering Mathematics

in the last two lectures I have
explained the elementary rules of
probability
so we firstly told that what are the
basic terminology of the which are used
in the study of probability for example
sample space events mutually exclusive
events disjoint events and so on then we
gave a eggs Matic framework which which
gives a theoretical foundation for the
study of probability theory then we saw
the rules for example addition rule the
theorem of total probability we give the
concept of conditional probability based
theorem and also the concept of
Independence of events today we will
elaborate all these concepts with the
help of several examples so let me start
with the first problem so two decks of
52 cards are mixed and well shuffled so
what is the meaning of mixed and well
shuffled that means when we draw the
cards the probability of drying each
card is same that means now when we are
having two decks there are 104 cards and
then if it is very shuffled then when we
draw randomly any card or when we
distribute the cards then each card will
have the same probability of occurrence
so that is a meaning of well shuffled
the cards are randomly distributed to
players to two players a and B each one
getting 52 cards so all the cards are
distributed so randomly we give one card
is given to first card for example from
the pack is given to one player second
card is given to one player so like that
and it is randomly distributed so both
the players get 52 cards what is the
probability that the player B does not
get a King of Hearts given that player a
gets it so let us analyze the event now
so in order to apply the theoretical
framework of our genetic definition we
define the events in the terms of the
sets so
let t be the event that the player a
gets a King of Hearts
now when two decks of cards are mixed up
there will be two kings of hearts so we
have to find out the conditional
probability that B does not get given
that a gets it so we define these events
what is the event E that the player a
gets a King of Hearts and similarly we
define the event F that the player B
gets a King of Hearts
so what is the required probability we
want that the player B does not get now
does not get means F complement given e
so this is the conditional probability
that is required so now you can see we a
problem which is a descriptive problem
in nature now we have reduced it to a
mathematical statement by making use of
the set theory here so the required
probability that the player B does not
get a king of Arts given that the player
a gets it is now simply the conditional
probability statement of F complement
given E so now we apply the definition
of the conditional probability this is
equal to probability of e intersection F
complement divided by probability of E
now the problem will be solved if we can
individually find out both of these
probabilities let us look at what is
probability of e intersection F
complement now if you see e is player a
gets the King of Hearts and F complement
means B does not get that means out of
the two King of Hearts both are with the
player a now you see what is the total
distribution out of 104 fifty-two cards
are distributed to the player a but out
of the 2 Kings he gets
he gets both the things that means out
of the hundred two cards see we can
write it like this to see - he gets both
the Kings and out of hundred two
remaining cards he gets the remaining 50
cards divided by 100 for c52 this is the
total number of ways of distributing 52
cards to the player a out of 104 cards
out of this he gets both the Kings so
two Kings were there he has selected two
and out of the remaining hundred two
cards he gets 50 cards so this is also
equal to 100 to see 52 divided by 104
see 52 now what is probability of E
again we can write it in terms of
probability of 1 minus probability of e
complement now what is a complement
ecomp lament again means that a does not
get a King of Hearts so it is the same
as B does not get a king of hearts
because there is nothing to distinguish
between a and B both are having
symmetric behavior here so this will be
since this probability is already
calculated probability of e complement
is same as probability of e intersection
F complement that means it will become 1
minus 100 to see 52 divided by 104 see
52 so substitute both of these terms in
this formula so we get probability of F
complement given E as after
simplification 51 divided by 150 5 which
is 0.32 9 or you can say point
approximately 0.33 that means there is a
33% or even say 1/3 of a chance that
player B does not get a King of Hearts
and given that a gets it let us look at
another problem
so there are three students let's call
them Sanjeev Prashant and Bharat they
have probability is point eight point
seven and point six two independently
solve a given problem that means they
are attempting the problem independently
and therefore whether one is able to
solve it or not has no effect on the
other so probability that Sangeeth will
be able to solve the problem is point
eight but Shawn will be able to solve
the problem is 0.7 the probability that
Bharat will be able to solve it is 0.6
now if the problem is solved what is the
probability that one only Sangeeth could
solve it our only Prashant could solve
it
our only God could solve it now this is
again the case of conditional
probability what is the probability that
Sangeeth could solve the problem given
that the problem is solved only Prashant
could solve the problem given that the
problem is solved etcetera so let us
again define let a denote the event that
Sangeeth could solve the problem B
denotes the event that Prashant could
solve the problem
let's see denote the event that Bharat
could solve the problem and let B be the
event that the problem is solved so what
we are interested is we are interested
in events of this nature probability of
a intersection B complement intersection
C complement given e because what does
this represent a represents Sangeeth
could solve it B complement will
represent that prashanth could not solve
it C complement would represent that
Bharat could not solve the problem given
that the problem is solved so let us
look at various probabilities here what
is probability of a probability of a is
0.8 probability of B is 0.7 probability
of C is 0.6 and what is probability of E
now he is that the problem is solved if
the problem is solved that means either
one of them could solve it or both of
our two of them could solve it are all
the thing could solve it so it is easy
to represent it as in the terms of
complimentary event 1 minus probability
complement that means the problem is not
solved if the problem is not solved that
is possible only if each of Sanjeet
prashanth and varrock could not solve it
that means e complement event can be
written as simultaneous occurrence of a
complement B complement and C complement
now here we have used some property of
the independence of events we have
assumed that these three students they
attempt the problem independently and
these probabilities are independent of
each other that means in terms of events
we can say that events a B and C are
independent now it can be proved that if
the events are independent then their
complements will also be independent so
this probability of the simultaneous
occurrence becomes the product of the
probabilities by the definition of
Independence so it is 1 minus
probability of a complement into
probability of B complement into
probability of C complement now once
again probability of ABC is given to a
so probability of complements can be
easily calculated by taking 1 minus so
you are getting 1 minus probability of a
complement will become 1 minus point 8
that is point 2 probability of e
complement becomes 1 minus point saving
that is point 3 probability of C
complement becomes 1 minus point 6 that
is 0.4 so after simplification this
becomes simply point nine seven six that
is the probability of e so probability
of a intersection B complement
intersection C complement given E that
is equal to probability of a
B intersection C intersection C
complement intersection e that will come
here divided by probability of e but
what is the event e event E is that the
problem is solved whereas a intersection
B complement intersection C complement
this is denoting that prashanth could
solve it Sangeeth could solve it but
prashanth invert could not solve it that
means ultimately the problem is solved
therefore this is a subset of e so we
can write it as simply probability of a
intersection B complement intersection C
complement only this term we need not
write here now this is point 8 once
again we use the independence here so
probability of B complement is 0.3 and
probability of C complement is 0.4
divided by probability of e so after
simplification the value turns out to be
6 by 61 now in a very similar manner we
can calculate the probability that
prashanth could solve it given that the
problem is solved so this will be
represented by probability of a
complement intersection B intersection C
complement given e and once again we
apply the same type of argument the
value turns out to be 7 by 122 and only
Bharath could solve it that will be a
complement intersection B complement
intersection C given E and this will
turn out to be 9 by 244 so this is an
application of the concept of
conditional probability as well as the
concept of Independence that we have
used here now let me take another simple
problem here now this is one of the
famous problems which is called birthday
problem here we are looking at a
modification of that same problem
suppose there are n persons in a party
and I am assuming the number of persons
is less than 12 what is the probability
that at least two have the same
birth month assume that each month is
equally likely now equally likely means
that we assume that each person has the
same probability of being born in a
given month although theoretically
speaking it need not be true because the
number of days in a month is different
but we make this assumption for
convenience here in this problem so if
there are say 13% then certainly two of
them will have the same birthday so here
we are taking the number of persons to
be less than 12 so let us assume a to be
the event that at least two have the
same birth month and a complement is the
event that no two have the same birth
month you have to find out the
probability that at least two have the
same birth month now in order to find
out this probability we have to
calculate exactly to have the same birth
month exactly three have the same birth
month and so on so this will be slightly
lengthy expression or you can say learn
the description of the event if we take
the complementary event it is much
simpler because we are simply saying
that each of them has a distinct birth
month so it will be easier to find out
the probability of a compliment in this
case you can see here probability of a
compliment there are 12 months and if n
distinct months have to be chosen from
here it will be done in twelve
p-n ways that is P denotes the
permutation here and the total number of
possibilities of birth months for n
persons will be 12 to the power n
because each person can have any of the
12 birth month so you will have 12 into
12 n times therefore probability of a is
simply 1 minus 12 P n divided by 12 to
the power n now this can also be written
in a slightly different form 12 into 11
into 10 and so on 12 minus n plus 1
divided by
12 to the power n which can also be
written as 1 - now this 12 by 12 112
will cancel out then you get it 11 by 12
we can write it as 1 minus 1 by 12 then
10 by 12 we can write as 1 minus 2 by 12
and so on 1 minus n minus 1 by 12 so
this is a nice representation of the
same thing now just to give an idea
about these values I have tabulated the
probability of a complement and
probability of a for different values of
n so you can see here then there are two
persons the probability that they have
the same birth month it is quite small
it's point 0 8 the probability of a
complement is 0.92 the probability of a
complement reduces as n increases and
probability of a increases as n
increases as you can see when there are
11 persons the value is almost 1 here in
fact you can see that after 7 persons
the value is that the probability is
almost 90% that at least 2 will have the
same birth month let us take another
example here now each of the
coefficients a B C in the system of
linear equations say a X plus B Y is
equal to zero + BX plus C Y is equal to
zero is determined by an independent
throw of a die and we assume that this
guy is fair okay so when we throw a die
a fair die you can get numbers 1 2 3 4 5
6 as the upper face each with the
probability 1 by 6 so once we throw it
whatever we the value we are associated
with a next time when we throw it
independently the number coming up is
associated with B and next time when we
throw it the number coming uppermost is
associated with C find the probability
that the system let me call it system
one has non-trivial solutions see this
system is a homogeneous system therefore
X is equal to zero Y is equal to zero
will always be a solution now
non-trivial solutions are possible if
the determinant of the coefficient
matrix is zero so non-trivial solutions
to one are possible if the rank of the
coefficient matrix is not 2 because if
this coefficient matrix is invertible
then you will have only one solution
that is 0 0 so if the determinant of the
coefficient matrix is 0 that is then AC
minus B square is equal to 0 now when we
toss the fair die three times and we get
the values ABC what are the
possibilities which will lead us to AC
minus B square is equal to 0 so you can
now see ABC each of them take value 1 2
3 4 5 6 what are the values of B Square
if B is 1 d square can be 1 now that
would give us only possibility that a
and C should also be 1 suppose I say B
is equal to 2 if B is equal to 2 then B
square is equal to 4 then if a and C are
to 2 then a into C will become 4
at the same time if a is equal to 1 C is
equal to 4 or a is equal to 4 C is equal
to 1 then also AC will become 4 that
means a is equal to B square similarly
if we look at say for example B is equal
to 3 then B square is equal to 9 then if
a and C are 3 then only AC will be equal
to 9 there is no other possibility and
similarly if we take B is equal to 4 B
square is equal to 16 and then you get
only possibility that a and C must also
be 4 in a similar way we can check all
the possibilities so the total number of
cases then you throw three times air
dies you will have 6 cube that is 2 1 6
and the favorable number of cases how
many that is ABC can take values 1 1 1 2
2 2 1 2 4 4 2 1 3 3 3 4 4 4 5 5 5 & 6 6
6 so there are total 8 cases so
probability of the system has
non-trivial solution
that will be equal to 8 by 2 1 6 that is
equal to 1 by 27 so there is a 1 in 27
chance that the system will have
non-trivial solutions in other cases the
system is going to have trivial
solutions here you observe we have made
the assumption that the dye is fair that
means each face has the same probability
of coming up the throws of the dyes are
considered to be independent and we have
just applied the classical definition of
probability for evaluation of this
particular probability let me give one
more application suppose there is a
tournament a chess player has to play
say three players and I am naming them
say top aloof Kramnik and Anand
once each the probabilities of
winning the game against them are so
against Sato Paulo let us assume that
probability is P one against say kramnik
suppose it is P 2 against Annan suppose
it is P 3 and these are things and
assume that p1 is greater than p2
greater than p3 the player let me call
it this player P the player P wins the
tournament if he or she wins two
consecutive games
I want to calculate that what is the
highest chance of winning the tournament
because there are can be various
possibilities he can play the players in
any order so he can choose the order in
which to play the three games let us say
what would be the best strategy for
winning the tournament that means in
which order he should play so that
winning the tournament is having the
highest probability
and another thing that we make
assumption that winning or losing in one
game should not have any effect on the
winning or losing of another game the
games are played independently I would
like to mention here that many of these
statements would not be written in the
classical books in the textbooks
many of these statements are not written
so when we are going to solve the
problem we have to put these assumptions
in the absence of these assumptions the
problem cannot be solved easily or it
may have altogether a different solution
now if there are three players let me
call them TKA then i can have six
permutations here that means I may play
t first then k and then a I may play t
first then a then K and so on so for
each of these ordering of playing the
games what are the probability of
winning the tournament so let us look at
this six possible orderings and their
probabilities of winning
- namond so let me list them consider
the first case you play T then K and
then a so denoting topple o Kramnik and
on now winning against top a low of
winning against so you have to win two
consecutive games that means you may be
in the first and second you may win
second and third however if you win the
first one and lose the second one and
then save in the third one even then you
don't win the tournament so you may say
what is the probability of that means
certainly I should win the second game
that is the probability of winning
against Kramnik is p2 so certainly this
should be possibility then out of T&A I
should win at least one that means if I
win all the three then also it's fine or
if I win t lose a or if I win a and
loose T then also it's fine so I can
write it as the probability of P 1 plus
P 3 minus P 1 into P 3 now what is this
probability coming this is probability
of t plus probability of a minus
probability of T intersection a that
means this is actually denoting
probability of T Union a so this is
probability of K so actually what is the
event K intersection T Union a and since
as a independence assumption is there
this becomes probability of K
in to probability of T Union a and
probability of T Union a among again
writing as probability of T plus
probability of a - probability of T into
probability of a because of independence
of T and a also so this is the term that
we will get now if we apply the same
argument we can consider all the
possibilities T a K so the second one we
should win that is P 3 and then out of T
and K at least one should be 1 so this
will become P 1 plus P 2 minus P 1 P 2
if we apply the same argument if you
look at the third we play K then T and
then a then you have to win the second
one that is P 1 then out of K and a you
should win at least one that is P 2 plus
P 3 minus P 2 P 3 then let us look at
for K ay T so you should certainly win
the second one that is P 3 then P 1 plus
P 2 minus P 1 P 2 then 5th option is a T
K that is equal to P 1 into P 2 plus P 3
minus P 1 minus P 2 P 3 and similarly
a K T that is P 2 into P 1 plus P 3
minus P 1 P 3 our original problem was
that what is the best strategy for
winning the tournament that means which
of these options give the highest
probability so we have to see that from
1 to 6 which which value is the highest
given that p1 is greater than p2 is
greater than P 3 now you observe here
the expressions here the expression 1 &
6 are the same the expression 2 & 4 are
the same expression 3 & 5 are the same
so basically we have to compare 1 2 & 3
so if you compare 1 2 & 3 note that 1 is
equal to 6 2 is equal to 4 3 is equal to
5 now you can see that 3 is greater than
1 why p1 is bigger than p2 now if you
compare these here because P 1 is bigger
than P 2 is bigger than P 3 then if you
look at this here we are getting P 1 P 2
that is cancelling out P 1 P 3 out of P
1 P 3 P 1 is bigger than P 2 so P 1 P 3
is bigger than P 2 P 3 the last term is
minus P 1 P 2 P 3 that is common
therefore 3 is bigger than 1 as P 1 is
bigger than P 2 similarly you can note
that 3 is bigger than 2 also why if you
multiply here you get P 1 P 2 this is P
1 P 3 so P 2 is bigger than P 3
therefore this term will be bigger
second term is P 1 P 3 and
here you are getting actually P 1 P 3 P
1 P 3 cancels out you get P 1 P 2 and
here you get P 2 P 3 so P 1 is bigger
than P 3 therefore this term will be
bigger and last term is P 1 P 2 P 3 so
since P 1 is bigger than P 3 3 is bigger
than 2 also so third option are the
fifth option now what is third option
third option is playing tupelov Anand
and Kramnik are playing onon I'm sorry
third is bigger so third is Kramnik
tupelo and onon and fifth is Anand -
Paulo and kramnik this is the best
strategy either 3rd or 5th now what is
the difference in the 3rd and 5th in
fact in both of them T is in the middle
and if you see the original
probabilities actually winning
probability against T is the highest
therefore if you play him second you
maximize your chances of winning so
playing
- Apollo second is the best strategy on
the other hand in fact if you play onon
in the middle then that is the worst
strategy because 2 4 is the smallest
value here because he is the strongest
player according to the probabilities of
winning against are given
a pair of dice is rolled until a sum of
seven are an even number appears what is
the probability that a seven appears
first so let us define the events here a
as the event that saving appears a pair
of dice so we are looking at the sum so
event a can happen as 1 6 2 5 3 4 4 3 5
2 are 6 1 so these are the possibilities
for a and B is the event let us say that
even number appears now even number
means the sum is even 1 1 1 3 so like if
you have the first one as 1 there are 3
possibilities if you have second first
one as 2 again there are 3 possibilities
like 2 2 2 4 2 6 and so on so there are
18 elements in B so probability of a
that will be equal to 6 by 36 that is 1
by 6 and probability of B will be equal
to 18 by 36 that is half also you can
see here that a and B are disjoint and
total number of elements in a union B
will be 24 that number of elements in a
union B that will be 24 so what is the
probability of a union B complement
there
we'll be equal to 12 by 36 that will be
equal to 1 by 3 so what is the
probability that 7 appears first now
this means that the event a occurs till
in the first throw itself so that means
the probability is 1 by 6 or in the
first throw 7 does not occur also even
number does not occur if that has going
to happen then the probability of that
is 1 by 3 and in the second throw 7
occurs or in the first 2 this does not
occur in the third one it occurs so you
get it as an infinite geometric series
the sum of this is simply equal to 1 by
6 divided by 1 minus 1 by 3 that is 2 by
3 so you get 1 by 4
I will give example of a famous matching
problem now matching problem is a very
classical problem in probability theory
it is like this so we have say for
example n envelopes and n letters are
written now these n letters are
addressed to n persons whose addresses
will be written on the envelopes however
when inserting the letters they are
randomly inserted so a famous problem is
that what is the probability that each
envelope receives the correct letter
into it now this problem as this is
known as matching problem it has various
versions for example there are n persons
say in males they have rings in their
hands with the mark of their say
would-be wives however when they
allocate the rings to the women they are
allocated randomly what is the
probability that all of them received
the correct ring similarly this problem
can be posed in various ways so we put a
abstract description so matching problem
I will call it n objects mark say 1 to n
they are distributed over n places which
are also marked 1 to N
that is one object is allocated to each
place what is the probability that
none of the objects occupies its correct
place that means one will not go to the
place one two will not go to the place
two three will not go to the place three
and so on what is the probability of it
now directly if we want to attempt this
problem this is slightly complicated
because if one is not going to one then
it can go to two three that means there
are n minus 1 possibilities if two does
not go to two then it can go to either
of the N minus 1 and so on so there will
be various possibilities however if we
look at the complementary event then it
is easier to analyze so we look at it in
this way
let us consider a to be the event that
none of the objects occupies its correct
place so a complement is the event that
at least one object occupies the correct
place then we can write a compliment as
union of AI I is equal to 1 to N where
AI denotes that the I ate object
occupies the I ate place for I is equal
to 1 to N then the event a compliment
that at least one object occupies the
correct place can be written as the
union of a is because this means at
least one of the events AI occurs at
least one means one may occur two may
occur and so on of course you can
observe that these a eyes are not
mutually exclusive a 1 a 2 a n they are
not mutually exclusive so in order to
evaluate probability of Union we will
apply the general addition rule now let
us look at this how the event AI occurs
if AI occurs when I yet place is
occupied by I ache object
and about other objects we are not
saying anything that means they may a Q
PI the correct place or they may not
occupy the correct place now remaining n
minus 1 objects are there and there are
N minus 1 places and each place will be
occupied by one object only this can be
done in n minus 1 factorial ways and
total number of ways of allotting n
objects 2 n places that will be n
factorial so what is the probability of
AI it will become simply n minus 1
factorial divided by n factorial and
remaining n minus 1 objects occupy any
of M minus 1 remaining that is occupy n
minus 1 remaining places in any fashion
now this can be done in n minus 1
factorial place so the probability of a
I becomes n minus 1 factorial divided by
n factorial which of course you can
write as 1 by n this will be for all ye
now if I want to calculate probability
of Union AI then I need probability of
AI is then I need probability of AI
intersection AJ I need probability of AI
intersection AJ intersection AJ and so
on so now you look at AI intersection AJ
how this will occur this means I yet and
J H places are occupied correctly
correctly means by IH + JH objects
respectively and remaining n minus two
objects occupy remaining n minus two
places in any order so probability of AI
intersection AJ that will become n minus
2 factorial divided by n factorial that
is equal to 1 by so we may just write it
like this itself now here I is not equal
to J of course we should take I less
than J because when we are writing down
the formula for the Union I need to take
only one order now continuing this
so if I write probability of AI
intersection AJ intersection a K that
will become n minus 3 factorial by n
factorial where I less than J less than
K and so on ultimately what is
probability of a 1 intersection a 2
intersection say a n now this means each
of the objects is in the correct place
that is the object number 1 object
number 2 object number n now this entire
thing can happen only in one way that
means each of them is going to unique
place so this probability will be simply
1 by n factorial
so if we apply the formula for
probability of a complement that is
probability of Union AI then that is
equal to Sigma probability of a I I is
equal to 1 to N minus double summation I
less than J probability of AI
intersection AJ plus probability of AI
intersection AJ intersection a K and so
on plus minus 1 to the power
plus one probability of intersection AI
is equal to 1 to M so now what is
probability of AI that was n minus 1
factorial by n factorial and there are n
terms so we write it as like NC 1 n
minus 1 factorial by n factorial now
these are NC 2 terms and probability of
AI intersection AJ was n minus 2
factorial by n factorial so this is n
minus 2 factorial by n factorial plus NC
3 these are n c3 and terms n minus 3
factorial by n factorial and so on and
this is simply one term so we can
simplify this this we can write as n
minus 1 n factorial divided by 1
factorial into n minus 1 factorial that
is simply 1 by 1 factorial if you look
at this this is n minus 2 sorry n
factorial divided by 2 factorial into n
minus 2 factorial so these terms cancel
out you get 1 by 2 factorial similarly 1
by 3 factorial and so on minus 1 to the
power n plus 1 1 by n factorial so
probability of a is nothing but 1 minus
this so 1 minus 1 by 1 factorial plus 1
by 2 factorial minus 1 by 3 factorial
minus 1 to the power N 1 by n factorial
in fact you can see that as n tends to
infinity this is nothing but e to the
power minus 1 so limit of probability a
as n becomes large is simply e to the
power minus 1 that is 0.36 7/8 that
means nearly
37% of the time the event that none of
the objects occupies the correct place
will be true which is quite high
probability I will give one more example
of the counting problems say football
clubs f1 and f2 are set to play a series
of say three games against each other to
decide the league champion the
probabilities of club f1 winning drying
and losing a game against f2 are half 1
by 8 and 3 by 8 respectively
that means f1 wins against f2 again with
probability 1/2 the probability of a dry
is 1 by 8 and the probability that f2
will win that is 3 by 8 that means f1
has slightly higher chance of winning
compared to F to a club gets 3 points
for a win one for a draw and 0 for a
loss so what is the probability that f1
wins our f2 wins are the league and in a
tie so we have to calculate the
probabilities of these events so here we
need to count all the cases so for
example let me make a tabular
representation here that will be easy to
understand what are the possibilities
let us count from the side of f1 say f1
we may win three games and it may not
lose any in that case it will have
points for f1 that will be nine and what
will be the points of f2
that will be zero so who will be winning
the league this will be f1 similarly if
you look at f1 wins - it draws one then
for two wins it gets six points for a
draw it gets one point so seven points
whereas for Wondra f2 gets a point
so the champion is still f1 like that we
count all the possibilities two wins one
loss
six points three points for f2 f1 will
win the league one win two draws it will
get five points f2 will get two points
and f1 will be the champion 5-1 win one
draw one loss in that case only four
points and this fellow will also get
four so the league will end in a tie let
us look at one win two losses then three
six f27 three draws three points for f1
3 points for f2 and ligands in a tie it
to draws one loss to five f2 1 draw two
losses one 7 F to 10 3 losses 0 9 f2 so
what are the possibilities for f1
willing the leaf for f1 winning the
league you have these cases what are the
probabilities here probability of f1
winning the league that is equal to
probability of these 1 2 3 4
possibilities probability of 1
probability of 2 probability of event 3
and probability of event for what are
the individual probabilities probability
of one is half cube because probability
of winning is half in the three games it
will have one half into half into half
so we are assuming the independence here
for the second one there are two losses
out of three games these can be chosen
in three see two ways the probability of
winning the to be MCS half square and
probability of a draw is won by eight
similarly in the third case 3c to half
square 3 by 8 plus 3c 1/2 + 1 by 8
square so this turns out to be 67 by 128
now similar way we can calculate what is
the probability of f2 winning the league
that turns out to be 170 1 by 5 1 2 and
probability of the league ending in ty
that will be equal to 70 3 by 5 1 2 so
we have given several examples of
solving probability problems using the
asthmatic definition in the following
lecture I will be explaining the concept
of random variables and probability
distributions