Probability
What is Probability
Many events can not be predicted for full certainty. For such events, we can say how
likely they are to occur.
The probability of an event is a measure of the likelihood that the event will occur.
If it is very certain that the event will occur, it's probability is 1 or close to 1
On the other hand, if the event is not likely to happen at all, it's probability is 0 or close to 0
For all other events, the probability ranges between 0 and 1.
In rainy season, when the sky is cloudy, the probability of receiving rains is very high.
However, in any other season, when the sky is clear, the probability of getting rains is very low.
The probability of event A is denoted by P(A).
It is computed as:
$$\text{Probability of an event happening P(A)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$$
Examples
Example 1
Suppose we draw a card from a deck of playing cards, what is the probability that the card is a King?
Solution: There are in all 52 cards in a deck. Out of this, 4 cards are Kings. If we can draw any of these 4 Kings,
the outcome is favourable to us.
$$\text{P(drawing a King)} = \frac{4}{52} = \frac{1}{13}$$
Example 2
Suppose we draw a card from a deck of playing cards, what is the probability that the card is a Heart?
Solution: There are in all 52 cards in a deck. Out of this, 13 cards are of Heart. If we can draw any of these 13 Heart cards,
the outcome is favourable to us.
$$\text{P(drawing a Heart)} = \frac{13}{52} = \frac{1}{4}$$
Example 3
Suppose we toss two coins together, what is the probability of getting 2 Heads?
Solution: There are 4 outcomes when we toss 2 coins together.
They are {(H,H), (H,T), (T,H), (T,T)}
Out of this, the favourable outcome for us is (H,H).
$$\text{P(getting 2 Heads)} = \frac{1}{4} $$
Terms used in Probability
Let us now look at the various terms used in the study of probability
 Experiment: The work or activity that generates the results to be studied.
 Outcome: The result of an experiment.
 Event: One or more outcomes of an experiment.
 Sample Space: A collection of all the possible outcomes of an experiment.
 Complement: The complement of an event is the event that does not happen.
 Independent Events: Two events are said to be independent when the outcome or occurrence of the first event
does not affect the outcome or occurrence of the second event.
When two events, A and B, are independent, the probability of both occurring is:
P(A and B) = P(A) . P(B)
 Dependent Events: Two events are said to be dependent when the outcome or occurrence of the first event
affects the outcome or occurrence of the second event.
Let us look at some examples to understand the above terms better.
Consider an experiment of rolling a dice.
Sample sapce in this experiment is {1,2,3,4,5,6}
In this experiment, if an event A is defined as getting an even number, then A = {2,4,6}
$$\text{P(even number)} = \frac{3}{6} = \frac{1}{2}$$
The complement of this event is
NOT getting an even number. The complement of an event A is denoted as A'.
A' = {1,3,5}
$$\text{P(NOT an even number)} = \frac{3}{6} = \frac{1}{2}$$
As a dice has numbers only upto 6, the probability of getting
7 as an outcome is not possible. It is called as an impossible event.
To understand the concept of independent events, let us look at some examples.
 • Getting a Heads after tossing a coin and drawing a King from a deck of cards.
 • Drawing a red marble from a jar and a getting an odd number on a spinner.
 • Receiving rain on a particular day and getting 4 on a throw of a dice.
The events in each of the above example are not related to each other and do not affect the outcome of one another.
A coin is tossed and a single 6sided die is rolled. Find the probability of landing on the head side of the coin and
rolling a 3 on the die.
$$\text{P(getting a Head)} = \frac{1}{2} $$
$$\text{P(getting a 3 on a die)} = \frac{1}{6} $$
$$\text{P(getting a Head and a 3 on a die)} = \frac{1}{2} . \frac{1}{6} = \frac{1}{12}$$
To understand the concept of dependent events, consider an experiment of cards.
A card is chosen at random from a deck of 52 playing cards. Without replacing the card, a second card is drawn.
What is the probability that the first card drawn is a King and the second card drawn is a Queen?
$$\text{P(getting a King in the first draw)} = \frac{4}{52} $$
$$\text{P(getting a Queen in the second draw given King is first draw)} = \frac{4}{51} $$
$$\text{P(getting a King and Queen)} = \frac{4}{52} . \frac{4}{51} = \frac{4}{663}$$
This leads us to a concept of
Conditional Probability.
Conditional Probability
The Conditional Probability of an event A in relationship to an event B is the probability that event A occurs given that event B
has already occurred. The notation for conditional probability is P(AB) This is read as the probability of event A given B.
When two events, A and B, are dependent, the probability of both occurring is
P(A and B) = P(A) . P(BA)
It is interesting to note that
P(A and B) = P(A) . P(BA) = P(B) . P(AB)
Example 1
Out of the 50 students surveyed in a school, 35 students play basketball in which 25 are boys. What is the probability that
if the student surveyed plays basketball then he is a boy?
Solution:
Let A be the event that the student is a basketball player and B be the event that the student is a boy.
Probability of the student being a basketball player, P(A) = $\frac{35}{50} $
Probability of the student being a basketball player and also a boy, P(A and B) = $\frac{25}{50} $
Probability of the student being a boy if he is a basketball player,
$$P(BA) = \frac{\text{P(A and B)}}{P(A)} =
\frac{\frac{25}{50}}{\frac{35}{50}} = \frac{25}{35} = \frac{5}{7} = 0.714$$
Example 2
In an exam, two math questions 1 and 2, are asked. 35% students solved problem 1 and 15% students solved both the problems.
How many students who solved the first problem will also solve the second one?
Solution:
Let A be the event that the students solved problem 1 and B be the event that the students solved problem 2.
Probability of the students solving problem 1, P(A) = 0.35
Probability of the students solving both the problems, P(A and B) = 0.15
Probability of students who solved the first problem will also solve the second one = $$P(BA) = \frac{\text{P(A and B)}}{P(A)}
= \frac{0.15}{0.35} = 0.429 $$
Example 3
The table below shows the math marks of a class of 50 students.
Find the probability that the student scoring full marks in a subject is a girl.


Less than Full marks



10



13

Solution:
Total boys in the class = 25
Total Girls in the class = 25
Probability of the student being a girl, P(Girl) = $\frac{25}{50}$
Probability of the student getting full marks and being a girl, P(Full marks and Girl) = $\frac{15}{50}$
Probability of Full marks scorer being selected being a girl = $$ \frac{\text{P(Full marks and Girl)}}{P(Girl)}
= \frac{\frac{15}{50}}{\frac{25}{50}} = \frac{15}{25} = 0.6$$