The document discusses the history and development of probability theory. It begins by describing how the foundations of probability were established in the 17th century by Blaise Pascal and Pierre de Fermat through their study of games of chance. It then provides definitions and explanations of key probability concepts like sample space, events, and the probability of an event occurring. The document goes on to discuss different models for representing probability problems geometrically, including linear models, area models, and coordinate geometry models. It also covers famous probability puzzles and paradoxes like the Monty Hall problem, Bertrand's paradox, and the Sleeping Beauty problem. Finally, it briefly mentions some applications of probability in fields like quantum mechanics, polymer sciences, and

1. By-
Harsha Hegde
Karthik N R
Manoj Hegde

2. A Time Travel into Probability
Two French mathematicians:
Blaise Pascal (1623–1662)
Pierre de Fermat (1601–1665) found the
mathematics of probability in the middle of the
seventeenth century.
Their first discoveries involved the probability of
games of chance with dice and playing card s .

3. BLAISE PASCAL (1623-1662)

4. PEIRRE DE FERMAT(1601-1665)

5. Introdoction to probability
Probability is ordinarily used to describe an attitude of
mind towards some proposition of whose truth we are not
certain.The certainty we adopt can be described in terms
of a numerical measure and this number, between 0 and 1,
we call probability.
The higher the probability of an event, the more certain
we are that the event will occur.
used widely in areas of study as mathematics, statistics,
finance, gambling, science, artificial intelligence/machine
learning and philosophy to, for example, draw inferences
about the expected frequency of events.

6. Probability of occurrence of an
event
P(n)=No of favourable outcomes
Sample Space(all possible outcomes)

7. Geometric Probability
Modern probability theory is full of formulas and
applications to modern events that are far removed from
games of chance. Instead of using algebraic formulas to
solve these probability problems, we use geometric
figures.
Geometric probability involves the distributions of length,
area, and volume for geometric objects under stated
conditions.
These geometric figures allow you to picture the
probabilities of a situation before solving the problem.
This will help you develop a sense of a reasonable solution
before you solve the problem.

8. Classical probability
Geometric Probability

9. Simple Problem Simulation
Imagine that you are throwing a dart at a 10x10 square
with a 4x4 square centrally placed. Then what is the
probability of you hitting the dart into the inner
block.?!

10. Favourable region=16 sq units
Feasible region=100 sq units
P= Area of favourable region =0.16
Feasible region
Thus you fancy a 16% chance of hitting the target!
This simple problem shows how logical is geometrical
probability compared to classical approach.

12. As is evident from the previous slide if the area of the
favorable region is increased consequently the
probability of the favorable event increases .Simple
but a very important conclusion!!

13. Various models in GP
Linear Models
Area Models
Co-ordinate geometry models
3D models

14. Linear Models
Example 1:
A is driving one car in a line of cars, with about 150
feet between successive cars. Each car is 13 feet long.
At the next overpass, there is a large icicle. The icicle
is about to crash down onto the highway. If the icicle
lands on or within 30 feet of the front of a car, it will
cause an accident. What is the chance that the icicle
will cause an accident?

16.  The chance that the icicle will strike a passing automobile may be represented
by line segments. The sample space is the distance from the front of one car to
the front of the next car. The feasible region is
A 163 feet B the 30 feet given in the problem plus the length of the car (13
feet).
Sample space Both of these may be represented as segments
Segment AB represents the sample space.
Sample space = 150 feet (distance between cars) + 13 feet (length of a car) = 163
feet
The feasible region may be represented by segment CD.
Measure of feasible region = length of segment CD = 43 feet

17. Probability=43/163=0.264
The chance that the icicle will cause an accident is
less than 30%!

18. Area Models
Example 2:The ARIZONA CRATER
A current scientific theory about the fate of dinosaurs
on the earth suggests that they were made extinct by
the effects of a large meteor which struck the earth in
the Caribbean Sea, off the coast of Central America.
The United States has been struck by large meteors in
the past, as shown by the crater in Arizona at left.
What is the chance that a meteor striking the earth
will land in the United States?

20. The probability relationships for this problem may be
represented by areas.
The sample space is the surface area of the earth, or
196,940,400 square miles.
The feasible region is the area of the United States, or
3,679,245 square miles.

21. Co-ordinate Geometry Models
Another model we can use to represent probability
problems is based on geometric figures placed on a
coordinate grid. The problem in this section
demonstrates how a coordinate system may be used
to solve probability problems.

22. Example
A line segment is 8 inches long. Two points are put
on the segment at random locations. What is the
probability that the three segments formed by the two
points will make a triangle?
This problem really is tedious and cumbersome in the
classical method..See for yourself how simplified it is
now!

24.  What is the probability that segment AB, segment BC, and
segment CD will make a triangle?
 This problem may be solved by using a coordinate system.
 let the length of segment AB = x. let the length of segment
BC= y.
 The length of segment CD = 8 - x - y.
 The length of AB must be less than 8 inches, and the
length of BC must be less than 8 inches. This may be
represented on a coordinate system as shown in Figure.

26. In addition, the sum of the two segments also must
be less than the total length of the original segment,
so AB + BC is less than 8 inches. This may be written x
+ y < 8. This inequality is shown on the coordinate
system in Figure .

28. The area of triangle QRS is the sample space. All the
points in triangle QRS represent possible lengths for
segment AB and segment CD. Point K, for example,
has coordinates (6, 1). Point K represents a length of 6
for segment AB, and a length of 1 for segment Be
The relationship between the three sides of a triangle
is as follows: The sum of the lengths of any two sides
of a triangle must be longer than the length of the
third side. This results in the following inequalities,
all of which must be true:

29.  x + y > 8 - x - Y (AB + BC> CD)
 x + (8 - x - y) > Y (AB + CD > BC)
y + (8 - x - y) > x (BC + CD > AB)

30. The first inequality may be
simplified as follows:
x+y>8-x-y
2x + 2y > 8
x + y> 4 (the shaded area in
Figure ).

31. The second inequality may
be simplified as follows:
x + (8 - x - y) > y
8> 2y
4> y, or y < 4 (the shaded
area shown in Figure ).

32. The third inequality may be
simplified as follows:
y + (8 - x - y) > x
8> 2x
4 > x, or x < 4 (the shaded
area shown in Figure ).

33.  The feasible region is the area of the
common region defined by all three
inequalities, as shown in Figure.
 The probability that the three
segments will make a triangle is
found by comparing the feasible
region to the sample space:
 Feasible region = Area of Triangle
EFG
 Sample space =Area of Triangle QRS

35. 3D models
Probability in 3D models is a problem rejoiced in
quantum physics. The normalization of the wave
function done over a 3D space is the application of
GP.
∫∫∫│w│2
dx dy dz
w-wave function

36. Buffon’s Needle Experiment
Georges Louis LeClerc, the Compte de Buffon
(1707–1788), was a mathematician who is now
famous for his needle experiment involving
probability.
Suppose we have a floor made of parallel strips
of wood, each the same width, and we drop a
needle onto the floor. What is the probability that the
needle will lie across a line between two strips?
He estimated the value of pi using this!(3.141592)

37. Bertrand’s paradox
The Bertrand paradox goes as follows: Consider an
equilateral triangle inscribed in a circle. Suppose
a chord of the circle is chosen at random. What is the
probability that the chord is longer than a side of the
triangle?
Bertrand gave three arguments, all apparently valid,
yet yielding different results.

38. The "random endpoints" method:
The probability that a random chord is
longer than a side of the inscribed
triangle is 1/3.

39. The "random midpoint" method
The probability a random chord is
longer than a side of the inscribed
triangle is 1/4.

40. The "random radius" method
The probability a random chord is
longer than a side of the inscribed
triangle is 1/2.

41. Monty Hall problem
The Monty Hall problem is a probability puzzle
loosely based on the American television game show
Let's Make a Deal and named after the show's original
host, Monty Hall.
It was originally posed in a letter by Steve Selvin to
the American Statistician in 1975 (Selvin 1975a) (Selvin
1975b). It became famous in the following form, as a
question from a reader's letter quoted in Marilyn vos
Savant's "Ask Marilyn" column in Parade magazine in
1990 (vos Savant 1990a):

42. Problem….
Suppose you're on a game show, and you're given the
choice of three doors: Behind one door is a car;
behind the others, goats. You pick a door, say No. 1,
and the host, who knows what's behind the doors,
opens another door, say No. 3, which has a goat. He
then says to you, "Do you want to pick door No. 2?" Is
it to your advantage to switch your choice?

43. In search of a new car, the player picks a door, say 1.
The game host then opens one of the other doors, say
3, to reveal a goat and offers to let the player pick door
2 instead of door 1.

44. behind door
1
behind door
2
behind door
3
result if
staying at
door #1
result if
switching to
the door
offered
Car Goat Goat Car Goat
Goat Car Goat Goat Car
Goat Goat Car Goat Car
The solution presented by vos Savant (1990b)
The probability of winning by staying with the initial choice is
therefore 1/3, while the probability of winning by switching is
2/3.

45. Player picks car
(probability 1/3)
Host reveals
either goat
Switching loses.
Player picks Goat A
(probability 1/3)
Host must
reveal Goat B
Switching wins
Player picks Goat B
(probability 1/3)
Host must
reveal Goat A
Switching wins

46. The player has an equal chance of initially selecting
the car, Goat A, or Goat B. Switching results in a win
2/3 of the time.

47. Two envelopes problem…
You have two indistinguishable envelopes that each
contain money. One contains twice as much as the
other. You may pick one envelope and keep the money
it contains. You pick at random, but before you open
the envelope, you are offered the chance to take the
other envelope instead.

48. Switching argument:
I denote by A the amount in my selected envelope.
The probability that A is the smaller amount is 1/2, and
that it is the larger amount is also 1/2.
The other envelope may contain either 2A or A/2.
If A is the smaller amount, then the other envelope
contains 2A.
If A is the larger amount, then the other envelope
contains A/2.
Thus the other envelope contains 2A with probability 1/2
and A/2 with probability 1/2.

49. So the expected value of the money in the other envelope is
This is greater than A, so I gain on average by swapping.
After the switch, I can denote that content by B and reason
in exactly the same manner as above.
 I will conclude that the most rational thing to do is to swap
back again.
To be rational, I will thus end up swapping envelopes
indefinitely.
As it seems more rational to open just any envelope than to
swap indefinitely, we have a contradiction.
The puzzle: The puzzle is to find the flaw in the very
compelling line of reasoning above.

50. Common resolutionobserve that A stands for different things at different places
in the expected value calculation, step 7 above
In the first term A is the smaller amount while in the second
term A is the larger amount
To mix different instances of a variable in the same formula
like this is said to be illegitimate, so step 7 is incorrect, and
this is the cause of
By definition in one envelope is twice as much as in the
other. Denoting the lower of the two amounts by X, we
write the expected value calculation as the paradox.
Here X stands for the same thing in every term of the equation

51. we learn that 1.5X is the average expected value in either of
the envelopes, hence there is no reason to swap the
envelopes.

52. Sleeping Beauty problem…
Sleeping Beauty volunteers to undergo the following
experiment and is told all of the following details: On
Sunday she will be put to sleep. Once or twice, during the
experiment, Beauty will be wakened, interviewed, and put
back to sleep with an amnesia-inducing drug that makes her
forget that awakening. A fair coin will be tossed to
determine which experimental procedure to undertake: if
the coin comes up heads, Beauty will be wakened and
interviewed on Monday only. If the coin comes up tails, she
will be wakened and interviewed on Monday and Tuesday.
In either case, she will be wakened on Wednesday without
interview and the experiment ends.

53. Any time Sleeping Beauty is wakened and
interviewed, she is asked, "What is your belief now for
the proposition that the coin landed heads?"

54. Solutions…
Thirder position:
The thirder position argues that the probability of heads is
1/3
Halfer position:
The halfer position argues that the probability of heads is
1/2
Phenomenalist position
The phenomenalist position argues that Sleeping Beauty's
credence is meaningless until it is attached to consequences

55. Applications
Quantum Mechanics
Polymer sciences
Ship wreck investigation
Deep sea divers
Meteorologists
Excavators