Probability
Probability is a very popular topic of the Mathematics syllabus of the IIT JEE. It is not considered to be a very simple topic but it is important to master the topic to remain competitive in IIT JEE.
In this chapter we shall study the methodologies of computing the probability of an event, be it complex or simple. We shall try to comprehend the tools used for this purpose. In an attempt to make the subject easily comprehensible and clearly understandable a large number of illustrations and problems are selected and solved. Every aspect is fully supported by sufficient number of relevant examples. The students will note that for the understanding of the subject matter of this chapter, the knowledge of permutation and combination (presented in the previous chapter) is essential.
Refer the following video for more information on Probability.
The syllabus includes:

Computation of probability of events using permutation and combination.

Addition and multiplication rules of probability,

conditional probability, independence of events.
The chapter starts with some basic definitions of

Experiments

Sample Space and

Event.
This is followed by the types of events.

Topics Covered under Probability are:

Experiment,Sample Space andEvent

Introduction to Probability

Conditional Probability

Independent events

Bayes’ Theorem

Binomial Distribution for successive events
Solved ExamplesProbability as the name suggests defines how probable an event may be. However, an event which is bound to occur is called certainty. It may also be defined as a measure or estimation of likelihood of occurrence of an event. Probabilities are given a value between 0 (0% chance or will not happen) and 1 (100% chance or will happen). The higher the degree of probability, the more likely the event is to happen, or, in a longer series of samples, the greater the number of times such event is expected to happen.
Some words have special meaning in Probability:
Experiment or Trial: A trial may be defined as an action where the result is uncertain. Some examples of trials include tossing a coin, throwing dice, drawing a card from a deck of cards.
Sample Space: All the possible outcomes of an experiment constitute the sample space. For instance if we consider the experiment of choosing a card from a deck of cards then the Sample Space is all 52 possible cards: {Ace of Hearts, 2 of Hearts, etc... }
The Sample Space is made up of Sample Points:
Sample Point: Just one of the possible outcomes is called a Sample Point.
Example: Deck of Cards
The 5 of Clubs is a sample pointThe King of Hearts is a sample point"King" is not a sample point. As there are 4 Kings that is 4 different sample points.
Event: A single result of an experiment is called an event. Some examples of events include:
Getting a Tail when tossing a coin is an eventRolling a "5" is an event.
We clarify the difference between these terms with the help of the following figure
Sample Space and Sample Point
It is clear from the above figure that while Sample space refers to the set of all possible outcomes; Sample Point is just one possible outcome. An event can be one or more of the possible outcomes.
You can refer the Past Year Papers to get an idea about the types of questions asked.
Independent Events: Two events are said to be independent of each other if the probability of the occurrence of one event does not affect the probability of the occurrence of the other event.
Example: Suppose you rolled a dice and tossed a coin. Now the probability of getting a particular umber on the dice in no way influences the probability of getting a head or a tail on the coin.
Conditional Probability:
The conditional probability of an event B is the probability that the event will occur given the knowledge that an event A has already occurred. This probability is written P (BA), notation for the probability of B given A. In the case where events A and B are independent (where event A has no effect on the probability of event B), the conditional probability of event B given event A is simply the probability of event B, that is P(B).
If events A and B are not independent, then the probability of the intersection of A and B (the probability that both events occur) is defined by P (A and B) = P (A) P (BA).
From this result, we also derive the following result
P (B\ A) = P ( A and B) / P (A)
Bayes’ Theorem: This theorem is based on the concept of conditional probability. Consider a finite set of mutually exclusive and exhaustive events Hi (i = 1, ..., n), i.e. events that satisfy
Hk ∩ Hm = Φ, for k ≠ m andH1 ∪ H2 ∪ ... ∪ Hn = Ω,
Then Bayes' theorem states that
P(HkA) = P(AHk) P(Hk) / ∑i P(AHi) P(Hi).
Binomial Distribution for Successive events: Binomial distribution is one of the most important distributions in the probability theory. The probability distribution that summarizes the likelihood that a value will take one of two independent values under a given set of parameters or assumptions is called the binomial distribution. The underlying assumptions of the binomial distribution are that there is only one outcome for each trial, that each trial has the same probability of success and that each trial is mutually exclusive.
Probability is not an easy topic but a tricky one. Despite of this it is not difficult to score in the questions based on this topic. The JEE exam definitely has 23 questions every year directly on these topics. It is very important to master these concepts as this becomes quite crucial in outperforming others in Competitive examination.
Illustration 1: A and B are two candidates seeking admission in IIT. The probability that A is selected is 0.5 and the probability that both A and B are selected is almost 0.3. Is it possible that the probability of B getting selected is 0.9?
Solution: Let P (A) and P (B) be the probabilities of selection of A and B respectively.
P (A) = 0.5, P (A∩B) ≤ 0.3
P(A∪B) = P(A) + P(B) – P(A∩B) ≤ 1
P(B) ≤ 1 + P(A∩B) – P(A)
≤ 1 + 0.3 – 0.5 ≤ 0.8
Hence, the probability of selection of B cannot be 0.9.
Illustration 2: A and B are two independent events. The probability that both A and B occur is 1/6 and the probability that neither of them occurs is 1/3. Find the probability of occurrence of A.
Solution: It is given that P (A).P (B) = 1/6
And also P (A^{c}) P (B^{c}) = 1/3
This can also be written as [1P (A)]. [1P (B)] = 1/3
Let P(A) = x and P(B) = y
Then (1x)(1y) = 1/3 and xy = 1/6
Hence, 1xy+xy = 1/3 and xy = 1/6
x+y = 5/6 and xy = 1/6
x(5/6 – x) = 1/6
6x^{2} – 5x + 1 = 0.
(3x 1) (2x1) = 0
So x = 1/3 and 1/2
Hence, the probability of A i.e. P(A) = 1/3 or 1/2.
Illustration 3: In a test an examinee either guesses or copies or knows the answer to a multiple choice question with four choices. The probability that he makes a guess is 1/3 and the probability that he copies the answer is 1/6. The probability that his answer is correct given that he copied it is 1/8. Find the probability that he knew the answer to the questions given that he correctly answered it.
Solution: let us use the following symbols for denoting the various options
G for guesses
C for copies
K denotes the possibility that the examinee knows
R if the answer is right
So, P(G) = 1/3
P(C) = 1/6
P(K) = 1 (1/3 + 1/6) =1/2
Now R = (R∩G) ∪ (R∩C) ∪ (R∩K)
P(R) = P(G) P(R/G) + P(C) P(R/C) + P(K) P(R/K) … (1)
Now, P(R/G) = 1/4
P(R/C) =1/8
P(R/K) =1
Putting this in equation (1), we obtain
P(R) = 1/3. 1/4 + 1/6.1/8 + 3/6. 1
= 1/12 +1/48 + 3/6
= 29/48
Hence, the required probability = P(K∩R)/ P(R)
= 24/29