Some Basic concepts of Probability along with advanced concepts on Medical probability & Probability in Gambling. A lot of Sample Questions and Practice Questions will help you understand and apply the concepts in real life.

1. Presented by - Dr.Prateek Goyal
Moderator – Dr. Namita Shrivastav

2.  Introduction
 Types
 Combining Probabilities
 Probability on a 2X2 TABLE
 Practice Questions

3.  A gambler's dispute in 1654 led to the creation of a
mathematical theory of probability by two famous French
mathematicians, Blaise Pascal and Pierre de Fermat. Antoine
Gombaud, a French nobleman with an interest in gaming and
gambling questions, called Pascal's attention to an apparent
contradiction concerning a popular dice game.

4. Probability is the branch of mathematics concerning numerical
descriptions of how likely an event is to occur. It can be of
majorly 3 types :
 Empirical or Experimental probability
 Theoretical or Classical probability
 Subjective Probability

5. Suppose we do an experiment of tossing a coin 1000 times in
which the frequencies of the outcomes are as follows:
Head : 455 Tail : 545
Based on this experiment, the empirical probability of a head is
455/1000, i.e., 0.455 and that of getting a tail is 0.545
These probabilities are based on the results of an actual
experiment of tossing a coin 1000 times. For this reason, they are
called Experimental or Empirical probabilities.
Moreover, these probabilities are only ‘estimates’. If we perform
the same experiment for another 1000 times, we may get
different data giving different probability estimates.
P(E) = Number of trials in which the event happened/
Total number of trials

6. But the requirement of repeating an experiment has some
limitations, as it may be very expensive or unfeasible in many
situations.
But If we are prepared to make an assumption of equally likely
outcomes, the repetition of an experiment can be avoided, as the
assumption will help us in directly calculating the exact
(Theoretical or Classical) probability.
P(E) = Number of outcomes favorable to E /
Number of all possible outcomes of the experiment

7. Now, let’s try to answer this question.
Q1 - What is the probability that Dr. Aritrik will top the upcoming
exam on Saturday?
Different people will have different opinions on this. Some may
say 90%, some may say 80% or some may say 100%.
This is what is known as Subjective Probability . It expresses one’s
degree of belief in a proposition.

8. Q2 - A bag contains a Red ball, a Blue ball and a Green ball, all the
balls being of the same size. Dr. Siddharth puts his hand inside the
bag without looking into it. What is the probability that what he
takes out is
A) A Red Ball
B) A Black Ball
C) A Ball

9. The probability of an event which is impossible to occur is 0. Such
an event is called an Impossible event.
The probability of an event which is sure (or certain) to occur is 1.
Such an event is called a Sure event or a Certain event.
Also, we have seen that the numerator (number of outcomes
favorable to the event E) is always less than or equal to the
denominator (the number of all possible outcomes). Therefore,
always, 0 ≤ P(E) ≤ 1

10. Q3 - Dr. Mohit tosses two different coins simultaneously. What is
the probability that he gets one head and one tail?
The possible outcomes are (H, H), (H, T), (T, H), (T, T), which are
all equally likely. Note that (H, T) & (T, H) are two different
outcomes.
The number of outcomes favorable to our Event is 2.
Therefore, P(E) = 2/4 = 50%

11. Q4 - Ganesh and Sangeeta are playing ludo. Ganesh’s gitti is 7
steps ahead of Sangeeta’s gitti. Two dice, one blue and one grey,
will be thrown at the same time by Sangeeta.
Sangeeta argues that there are 11 possible outcomes - 2, 3, 4, 5,
6, 7, 8, 9, 10, 11 and 12. Therefore, probability of Ganesh’s gitti
being cut is only 1/11.
What is your stand on this?

12. All the possible outcomes are:
1,1 2,1 3,1 4,1 5,1 6,1
1,2 2,2 3,2 4,2 5,2 6,2
1,3 2,3 3,3 4,3 5,3 6,3
1,4 2,4 3,4 4,4 5,4 6,4
1,5 2,5 3,5 4,5 5,5 6,5
1,6 2,6 3,6 4,6 5,6 6,6
There are 6 favourable outcomes for our event. So P(E) = 6/36 =
1/6 and not 1/11
Also, we see, that the probability of getting a double 6 is 1/36

13. There are many situations in which the outcome is some point
within a circle or rectangle, etc.
But Can we count the number of all possible outcomes in such
case?
We know, this is not possible since there are infinitely many points
within a circle. We deal with such problems in a different way.

14. Q5 – A helicopter crashed in the area shown in the Figure. What is
the probability that the helicopter is inside the lake shown in the
figure ?

15. The helicopter is equally likely to crash anywhere in the region.
Area of the entire region where the helicopter can crash = (4.5 ×
9) km2 = 40.5 km2
Area of the lake = (2.5 × 3) km2 = 7.5 km2
Therefore, P (Helicopter crashed in the lake) = 7.5/40.5 = 18.5 %

16. Q6 - Complete the following statements:
(i) The probability of an event that cannot happen is ______. Such
an event is called _______.
(ii) The probability of an event that is certain to happen is ______.
Such an event is called _______.
(iii) The sum of the probabilities of all the outcomes of an
experiment is _________.

17. Q7a - A box contains 12 balls out of which x are black. If one ball is
drawn at random from the box, what is the probability that it will
be a black ball?
Q7b - If 6 more black balls are put in the box, the probability of
drawing a black ball is now double of what it was before. How
many black balls were originally there in the box?
Probability 1 = x/12
Probability 2 = x+6/12+6
Also given that Probablity2 = 2 X Probability 1
So, x+6 / 18 = 2 x/12
X comes out to be 3. So originally, there were 3 balls.

18. Q8 - A box contains 5 red marbles, 8 white marbles and 4 green
marbles. One marble is taken out of the box at random. What is
the probability that the marble taken out will be not green?
Probability (Green) = 4/17
Probability (Not Green) = 1-4/17 = 13/17
P (not E) = 1 - P(E)
P (not E) is also represented as P(Ē) and is known as the
complement of P(E)

19. Combining
Probabilities

20. Q9 - A box contains 5 red marbles, 8 white marbles and 4 green
marbles. One marble is taken out of the box at random. What is
the probability that the marble taken out will be red or green?
Probability (Red) = 5/17
Probability (Green) = 4/17
Probability (Red or Green) = 5/17 + 4/17 = 9/17
This is known as the additive rule of probability. It is valid
when both the events are mutually exclusive, i.e, only one
event out of the desired events can occur at a time.
P(A or B) = P(A) + P(B)

21. Q10 – What is the probability of getting a King or a hearts
when a card is randomly picked from a deck of 52 cards?
Probability (King) = 4/52
Probability (Hearts) = 13/52
Probability (King or Hearts) = 4/52 + 13/52 = 17/52
But this is not true as the problem here is both events are not
mutually exclusive. A king of hearts is the problem here.
So in such situations, additive rule is something like:
P(A or B) = P(A) + P(B) - P(A&B)
So Solution will be 4/52 + 13/52 – 1/52 = 16/52

22. Q11 - Suppose now a marble is drawn and returned to the bag
and then a second marble is drawn. What is the probability of
obtaining a red marble first and then a green one?
(5 red marbles, 8 white marbles and 4 green marbles)
In this situation, The Multiplicative rule of probability holds and the
probability of joint occurrence of two independent events is given
by the multiplication of the separate probabilities.
P (A & B) = P (A) X P (B)
So answer is 5/17 X 4/17 = 20/289= 0.07 Approx
The requirement of independence means that occurrence of the
first event does not affect the probability of the second event. If the
result of the first draw can have an influence on the result of the
subsequent draw, a different rule must be used.

23. Q12 - Suppose now a marble is drawn but not returned to the bag
and then a second marble is drawn. What is the probability of
obtaining a red marble first and then a green one?
(5 red marbles, 8 white marbles and 4 green marbles)
Probability (Red) = 5/17
Probability (Green) = 4/16 (not 4/17)
P (Red & Green) = 5/17 x 4/16 = 0.735%
The formula is
P(A & B) = P(A) X P(B given A)
This is also known as conditional probability.
P(B given A) is represented as P (BǀA)

24. Q13a – What is the probability of a couple having first three
children as girls?
Probability of first child being a girl = 0.5
Probability of second child being a girl = 0.5
Probability of third child being a girl = 0.5
Probability of first three child children being girls = 0.5 X 0.5 X 0.5=
0.125
Q13b – What is the probability of a couple having their 4th child as
a boy after they have already had 3 girls?
0.5

25. Q 14 a - Suppose a bag contains 4 red, 3 purple and 2 pink balls.
What is the probability that first draw is a purple ball?
Probability (Purple Ball) = 3/9 = 0.33%
Q14b – We didn’t put the ball back. What is the probability that a
second draw will reveal a purple marble?
Probability (Purple Marble) = 0 % ( There are no marbles in bag)

26. Applying Probability on a 2X2 table

27. True Disease State
D+ D-
Test
Results
T+ 7 4 11
T- 3 86 89
10 90 100

28. True Disease State
D+ D-
Test
Results
T+ 7 4 11
T- 3 86 89
10 90 100
Q1 )What is the probability that a person selected at random
will have the disease as determined by
1) True disease state
2) Screening test
– 10/100
– 11/100

29. True Disease State
D+ D-
Test
Results
T+ 7 4 11
T- 3 86 89
10 90 100
Q2) What is the probability that a person who has the disease
will also tests positive on the screening test?
Persons who have the disease = 10
Person who tested positive = 7
Probability = 7/10
This is also known as sensitivity of the test. (True Positives /
Total Diseased)
Can be represented as P(T+ǀD+)

30. True Disease State
D+ D-
Test
Results
T+ 7 4 11
T- 3 86 89
10 90 100
Q3) What is the probability that a person who does not have the
disease will also tests negative on the screening test?
Persons who do not have the disease = 90
Person who tested negative = 86
Probability = 86/90
This is also known as specificity of the test. (True Negatives
/Total Non-Diseased)
Can be represented as P(T-ǀD-)

31. True Disease State
D+ D-
Test
Results
T+ 7 4 11
T- 3 86 89
10 90 100
Q4) What is the probability that a person who has tested positive
for the disease on the screening test will have the disease?
Persons who have tested positive for the disease = 11
Person who actually are diseased = 7
Probability = 7/11
This is also known as Positive Predictive Value of the test.
(True Positives /Total Positives)
Can be represented as P(D+ǀT+)

32. True Disease State
D+ D-
Test
Results
T+ 7 4 11
T- 3 86 89
10 90 100
Q5) What is the probability that a person who has tested
negative for the disease on the screening test will not have the
disease?
Persons who have tested negative for the disease = 89
Person who actually are not diseased = 86
Probability = 86/89
This is also known as Negative Predictive Value of the test.
(True Negatives /Total Negatives )
Can be represented as P(D-ǀT-)

33. Practice Questions

34. Q1) What is the probability of me getting a 6 three times
consecutively in a game of Ludo?
Q2) A jar contains 24 marbles, some are green and others are
blue. If a marble is drawn at random from the jar, the
probability that it is green is 2/3. Find the number of blue
balls in the jar.
Q3) What is the probability of me loosing weight in the
upcoming month?
Q4)What is the probability of me getting a spade if I randomly
pick a card from a deck of 52 cards and then throwing an odd
number on a dice?

35.  Q5) Two PG’s Dr. Glory and Dr. Alam are asked to discuss their
protocol with their guide only once in a week, only after lunch
break. What is the probability that both will visit their guide
on (i) the same day? (ii) consecutive days?
G- Mon
A-Mon
G- Tue
A-Mon
G- Wed
A-Mon
G- Thu
A-Mon
G- Fri
A-Mon
G-Mon
A-Tue
G-Tue
A-Tue
G-Wed
A-Tue
G-Thu
A-Tue
G-Fri
A-Tue
G-Mon
A-Wed
G-Tue
A-Wed
G-Wed
A-Wed
G-Thu
A-Wed
G-Fri
A-Wed
G-Mon
A-Thu
G-Tue
A-Thu
G-Wed
A-Thu
G-Thu
A-Thu
G-Fri
A-Thu
G-Mon
A-Fri
G-Tue
A-Fri
G-Wed
A-Fri
G-Thu
A-Fri
G-Fri
A-Fri
Same Day = 5/25 = 20%
Consecutive Days = 8/25 =32%

36. Q6) A Senior Resident has given a problem individually to 5 first
year PGs. The chances of solving it in one day are:
 Sunom -1/2
 Siddharth - 1/3,
 Chirag- 1/4,
 Ravindra - 1/5
 Vignitha - 1/6
What is the probability that the SR will receive the solution by
any one of them next day?

37. First, we need to find out the probability of all of them failing to
find a solution. In that case only, SR will not receive a solution.
Probability that Sunom fails to solve the problem is 1-1/2 =1/2
Probability that Siddharth fails to solve the problem is 1-1/3=2/3
Probability that Chirag fails to solve the problem is 1-1/4=3/4
Probability that Ravindra fails to solve the problem is 1-1/5=4/5
Probability that Vignitha fails to solve the problem is 1-1/6=5/6
Since the events are independent the probability that all the five
first year PGs fail to solve the problem is :
1/2 ×2/3×3/4×4/5×5/6 =1/6
The probability that the problem will be solved
1-1/6 =5/6

38. True Disease State
D+ D-
Test
Results
T+ 80 50 130
T- 20 850 870
100 900 1000
Q7) If an individual from this population tests positive on the
screening test, what is the probability that the individual has the
disease? What is this known as? How can it be represented?
80/130 ,PPV
Represented as P(D+ǀT+)

39. True Disease State
D+ D-
Test
Results
T+ 80 50 130
T- 20 850 870
100 900 1000
Q8) If an individual from this population is known to be disease
free, what is the probability that the individual tests positive on
the screening test? How can it be represented?
50/900
Represented as P(T+ǀD-)

40. Q9 a)What is the probability of a 6 turning up once in 3 times
when a dice is rolled. (Think Hard)
1. There are 6 sides to a die, so there is 1/6 probability for a
6 to turn up in 1 throw.
2. That is, there is 1−1/6=5/6 probability for a 6 not to turn
up.
3. When you throw a die 3 times, the probability of a 6 not
turning up at all is (5/6)3. Which equals to 0.58. So
probability of a 6 turning up is 0.42
Q9 b)What is the probability of a 6 turning up once in 6 times
when a dice is rolled.
1- (5/6)6 = 0.67
This is a known as veridical paradox. Probability is different from
what it would be expected in a simpler way.

41. Q10) Now in the end, let’s try to answer the problem raised by
Antoine Gambud to Blaise Pascal . Which of these two is more
probable:
Getting at least one six with four throws of a die
or
Getting at least one double six with 24 throws of a pair of dice?
Also, when rolling the dice 24 times to get a double 6, it looked like
he has chances of winning 2 out of 3 times. But this was not
happening.

42. 1. There are 6 sides to a die, so there is 1/6 probability for a 6 to
turn up in 1 throw.
2. That is, there is 1−1/6=5/6 probability for a 6 not to turn up.
3. When you throw a die 4 times, the probability of a 6 not
turning up at all is (5/6)4. Which equals to 0.48. So probability
of a 6 turning up is 0.52
1. Similarly, when we throw two die, there is 1/36 probability of
double 6 turning up in 1 throw.
2. That is, there is 1−1/36=35/36 probability for a 6 not to turn
up.
3. When you throw a die 24 times, the probability of a 6 not
turning up at all is (35/36)24. Which equals to 0.51. So
probability of a 6 turning up is 0.49