Probability is quite a complex chapter, yet an important one in JEE Main Mathematics. The questions asked from this section are usually very scoring. Around 2-3 questions are asked from Probability which bear a total of about 8 marks. Therefore, the weightage of this chapter in JEE Main is 2-3%. Check JEE Main 2020 Mathematics Syllabus

• Some of the topics tested under Probability are Conditional Probability, Theorem of Total Probability, Independent Events, Bayes' Theorem, and Bernoulli Trails.
• NCERT books are a great resource for understanding these concepts. A lot of consistent practice is required to ace this chapter.

Must Read:

• JEE Main Study Notes for Differential Calculus
• JEE Main Study Notes for Integral Calculus

Read this article for study notes on Probability which will help you in JEE Main preparation. 

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What is Probability?

In the easiest possible words , the degree of likeliness of an event occurring is probability. Predicting an event with complete certainty is not always possible. The definition of probability is used in these situations to predict how likely an event is to happen or not. It may also be considered as a measure or estimate of the likelihood of an event happening. Probabilities are given a value from 0 (0 percent chance or not) to 1 (100 percent chance or will happen). The greater the degree of probability, the more likely the event is to occur, or the higher the number of times such an event is expected to take place in a lengthier series of samples.

• N(A) denotes the number of favourable outcomes to A and n(S) denotes the total number of outcomes in sample space. Thus, the formula above becomes P(A) = n(A)/n(S). 
• The probability of an occurrence may vary from 0 to 1, that is, 0 to p 1. 
• Probability cannot ever be negative. 
• Probability of occurrence of a case = 1- (Probability of not occurring).

With the help of the following examples, probability can be explained better - 

1. Throwing a dice: Throwing a dice will produce six potential outcomes. We can get any number between 1 and 6, and therefore the total number of potential outcomes is 6. Therefore, the probability of being given any number between 1 to 6 is 1/6. But what is the likelihood of getting a 7 over a dice throw? Since it is not feasible to get a 7 on a dice, its likelihood is nil. 

2. Tossing a coin: We may either get a head or a tail when a coin is tossed. So, two possible outcomes can be achieved, and the probability of each is 1⁄2. 

Therefore, P(H) = P(T) = 1⁄2

Experiment or Trial

An experiment or a trial means an action whose result is uncertain. It is an operation that could produce certain well defined outcomes. A few examples of trials could include throwing dice, tossing a coin, and drawing a card from a deck of cards.

The two types of experiments are - 

• Deterministic - Those experiments, which when repeated under similar conditions, give the same outcome are called deterministic. This kind has only one possible outcome and the results are absolute and unique. For example, an experiment conducted for verification of Newton's Law of Motion. 
• Random or Probabilistic - Those experiments, which when repeated under similar conditions, do not give the same outcome every time are called random. The outcomes of these experiments are known beforehand without any specific outcome. For example, throwing a dice is a random experiment because it is a well-known fact that there are only six possible outcomes. However, it is not possible to know which one will be the result.

Sample Space

Sample space refers to all the possible outcomes of an experiment. For example, there are 52 cards in a deck. Therefore, the sample space is all of the 52 cards. 

Sample Point

A sample space consists of sample points that refers to only one of the probable outcomes. 

For example, in a deck of cards, obtaining the King of Hearts is a sample point. Note that the "King" is not a sample point because there are 4 Kings that contain 4 distinct sample points. Another example is obtaining a ‘2’ on throwing a dice. 

Event

An event is a single result of an experiment. A few examples of events could include getting a tail when you toss a coin.

Types of Events

1. Elementary Event: It is an outcome of a random experiment.

For example, consider the random experiment of tossing a coin. The probable outcomes of this experiment are head (H) or tail (T).

Thus,

E1 = Getting head (H) on the upper face of the coin,

and

E2 = Getting tail (T) on the upper face of the coin.

Then, E1 and E2 are elementary events that are associated with the experiment

2. Compound Event: This is when an event is received from a random experiment by combining two or more elementary events that are associated with the random experiment.

For example, in one throw of a dice, the event of “getting an even number” is a compound event that is received through combining elementary events 1, 3, and 5.

3. Certain Event: It is an event that consists of all the sample points of sample space.

For example, a random experiment of a card that is drawn at random from a deck of 52 cards. Assume event A is that the card drawn is either red or black. Therefore, event A is a certain event

4. Impossible Event: It is an event that does not consist of any sample point of the sample space S. Null set Ø expresses an impossible event.

For example, let event B be that the dice shows a number higher than 6. Therefore, B= {} = Ø

Complement of an Event

Assume that A is an event of sample space S. The set of all outcomes that are in S but not in A is known as the complement of the event A and is expressed as A’.

Mutually Exclusive Events (Disjoint Events)

Assume that S is the sample space. A and B are two events of S. Events A and B are to be mutually exclusive if they do not share any sample point.

Independent Events

Two events are considered to be independent of each other when the probability of the happening of one event has no effect on the probability of the occurrence of the other event.

For example, suppose you tossed a coin and rolled a dice. The probability of getting a specific number on the dice does not in any way influence the probability of getting a head or a tail on the coin. 

Note: 

• Three events A, B and C are independent when the following conditions hold good:
• P(A ∩ B) = P(A) . P(B), P(B ∩ C) = P(B) . P(C), P(C ∩ A) = P(C) . P(A)
• P(A ∩ B ∩ C) = P(A).P(B).P(C) 
• Therefore, three events are considered to be independent if they are independent in pairs and mutually independent.
• In a similar manner, when we have n events A1, A2, …. An, they are said to be independent when they satisfy nC2 + nC3 + …. + nCn = 2n – n – 1 conditions. 

Probability Distribution

Probability Distribution is a table or an equation which connects every outcome of a statistical experiment to its probability of occurrence.

In order to fully understand probability distribution, it is essential to understand variables, random variables, and notation.

• A variable is a symbol ( A,B , x, y, and so on) that could pick up any specified set of values. 
• If the value of a variable is the outcome of the statistical experiment, that specific variable is a random variable:
  • X expresses the random variable X.
  • P(X) expresses the probability of X.
  • P(X=x) implies the probability that the random variable X is equal to a particular value represented by x.

If a coin is tossed 2 times, the statistical experiment could have 4 possible outcomes: HH, HT, TH, and TT. Assume that variable X expresses the number of Heads which result from the experiment. The variable X could pick up the values 0, 1 or 2. In this case, X is a random variable.

The table given below depicts every outcome that is associated with its probability. It holds reference to the experiment mentioned above with X as the number of heads.

Conditional Probability

The probability of a specific event can sometimes depend on the occurrence or non-occurrence of another event.

Assume that A and B are two events in a sample space S.

Let n = number of sample points in S

m1 = number of sample points in A

m2 = number of sample points in B

m12 = number of sample points in A∩B.

Then,

P(A) = m1 / n , P(B) = m2 /n and P(A∩B) = m12 /n

The probability of A∩B, (i.e. of B) in the sample space A is m12 / m1. It is the probability of B that comes under the assumption that A occurs. It is represented by P(B/A) and is known as the conditional probability of B only when A happens. 

Thus, P(B/A) = m12 / m1 = n(A∩B)/n(A), if n(A) ≠ 0.

In a similar manner, P(A/B) = m12 / m2 = n(A∩B)/n(B), if n(B) ≠ 0.

The two events A and B are considered to be independent, if P(A/B)=P(A) and P(B/A) = P(B).

Bayes Theorem

Bayes Theorem explains the probability of an event. This explanation is based on the prior knowledge of situations that could be associated with the event.

For example, cancer is associated with age. By using Bayes theorem, a person’s age could be used to assess the probability that he/she has cancer. This is more accurate than evaluating the probability of cancer without the presence of prior knowledge of the person’s age.

The probability mentioned under Bayes theorem is also called by the name of inverse probability, posterior probability, or revised probability. It finds the probability of an event through consideration of the given sample information. Hence the name posterior probability. The Bayes theorem is founded on the formula of conditional probability.

If A1, A2, A3,…..An are n mutually exclusive events in the sample space S, B is any other event from S. If the probability of occurrence of Ai’s and probability of occurrence of B if Ai, i=1,2,3,…n has happened are known, then the probabilities of occurrence of Ai’s if B has happened are expressed as -

Binomial Distribution for Successive Events

Binomial Distribution is one among the most important distributions in probability theory. It is the probability distribution that summarizes the likelihood that a value will pick up one of two independent values within a specified set of parameters or assumptions.

Assume that the probability of success in any trial is p and that of failure is q. Then, 

p + q = 1

(p + q)n = C0 Pn + C1Pn-1q +...... Crpn-rqr +...+ Cnqn

Therefore, the probability of exactly k successes in n trials is represented by -

Pk = nCkqn-kpk

Assumptions of Binomial Distribution

1. The number of observations n is fixed.
2. It provides one of two probable outcomes. For example, success when the required event occurs and failure when the required event does not occur.
3. Every trial has the same probability of success.
4. Every one of the trial is mutually exclusive.
5. The probability of getting at least k successes is P(x > k) = Σnx=k nCx px qn-x
6. Σnx=k nCx qn-x px = (q + p)n = 1.
7. The mean of binomial distribution is np.
8. The variance is npq. 
9. The standard deviation is represented by (npq)1/2, where n,p, and q derive their meanings as described previously. 

Points to Remember while Solving Questions on Probability

• The sum of binomials is also binomial. If X ~ B(n, p) and Y ~ B(m, p) are independent binomial variables having the same probability p, then X + Y is also a binomial variable with the distribution X + Y ~ B(n + m, p).
• When X ~ B(n, p) and conditional on X, Y ~ B(X, q), then Y is a simple binomial variable having the distributionY ~ B( n, pq).
• The Bernoulli distribution is a unique scenario of the binomial distribution, where n = 1. Symbolically, X ~ B (1, p) holds a meaning similar to X ~ Bern (p).
• Suppose an experiment contains only two possible outcomes, it is said to be a Bernoulli trial. The two outcomes are success and failure.
• Any binomial distribution, B (n, p), is the allotment of the sum of n independent Bernoulli trials Bern (p), each of them having the same probability p.
• The binomial distribution is a unique scenario of the Poisson Binomial Distribution which is a sum of n independent non-identical Bernoulli trials Bern(pi). When X has the Poisson binomial distribution with p1 = … = pn = p, then X ~ B(n, p).
• A cumulative binomial probability implies the probability that the binomial random variable lies amidst a given range (For example, is greater than or equal to a specified lower limit and less than or equal to a specified upper limit).
• A negative binomial experiment is an experiment that contains x repeated trials in which every trial could result in only two possible outcomes - a success or a failure. Along with this, it has the properties mentioned below -

1. The probability of success, represented by P, is the same on each trial.

2. The trials are independent. The outcome on one trial has no effect on the outcome on other trials.

3. The experiment goes on until r successes are seen, where r is mentioned in advance.

• Assume that A1, A2, .... An is a set of mutually exclusive and exhaustive events and E is an event that is linked to A1, A2, ...., An. Then, the probability that E occurs is expressed as -
• P (E) = ∑n(i=1) P(Ai)P(E/Ai).
• When the set of n events associated with a sample space are independent pairwise, they should be mutually independent, but the vice versa is not true all the time.
• In probability questions that require the application of the total probability formula, the events A1 and A2 should satisfy the three conditions specified below - 

1. A1 ∩ A2 = Φ

2. A1 U A2 = S.

3. A1 ∈ S and A2 ∈ S. 

• In order to identify the questions in which the Bayes’ theorem has to be used, the key phrase to look for “is found to be".

Previous Year Solved Questions

Ques. Suppose A and B are two candidates seeking admission in IIT, the probability that A is chosen is 0.5 and the probability that both A and B are chosen is almost 0.3. Is it probable that the possibility of B getting chosen 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

Thus, the probability of selection of B cannot be 0.9.

Ques. Assume that A and B are two independent events. The probability that both A and B will happen is 1/6 and the probability that neither of them will happen is 1/3. Calculate the probability of A’s occurence.

Solution: Given, P (A).P (B) = 1/6

P (Ac) P (Bc) = 1/3

It can even be written as [1-P (A)]. [1-P (B)] = 1/3

Let P(A) = x and P(B) = y

Then, (1-x)(1-y) = 1/3 and xy = 1/6

Hence, 1-x-y+xy = 1/3 and xy = 1/6

x+y = 5/6 and xy = 1/6

x(5/6 – x) = 1/6

6x2 – 5x + 1 = 0.

(3x -1) (2x-1) = 0

So x = 1/3 and 1/2

Thus, the probability of A, that is, P(A) = 1/3 or ½.

Ques. In an examination, a candidate either guesses, copies, or knows the answer to an MCQ with four choices. The probability that he guesses is 1/3 and the probability that he copies is 1/6. The probability that his answer is correct if he copies is 1/8. Find the probability that he knows the answer to the questions provided that he correctly answered it.

Solution: Symbols used -

G for guesses

C for copies

K expresses 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), 

P(R) = 1/3. 1/4 + 1/6.1/8 + 3/6. 1

= 1/12 +1/48 + 3/6

= 29/48

Thus, the required probability = P(K∩R)/ P(R)

= 24/29

Ques. A number x is chosen from the first 100 natural numbers. Find the probability that x meets the condition x + 100/x > 50.

Solution: The total number of ways of choosing x is 100.

Given, x + 100/x > 50.

This equation is satisfied for all the numbers x such that x > 48 and also for x = 1 and 2

Therefore, the favourable number of cases is 55.

Thus, probability = 55/100 = 11/20.

Ques. There are three events A, B and C. Only one of them must and can happen. The odds are 8 to 3 against A and 2 to 5 for B. Find the odds against C.

Solution: P(A) = 3/11, P(B) = 2/7, P(C) = x (say)

Because only one event can happen, A, B, and C are exhaustive and mutually exclusive events.

So, P(A) + P(B) + P(C) = 1

⇒ 3/11 + 2/7 + x = 1

x = (77 – 21 – 22)/77 = 34/77

Odds against C = (77 - 34) : 34 or 43 : 34

Tricks to Solve Questions from Probability

1. Solve a minimum of 20 Probability questions while preparing for the chapter. 
2. Ask your mentors to help you out with Probability questions whenever you are stuck.
3. Use Goyal’s “A Textbook of Algebra for JEE Mains and Advanced” by Arihant Publishers and IIT Mathematics by M.L.Khanna for better practice. 
4. Attempt as many mock tests as you can because you can Probability only with consistent trying. 
5. Solve the problems using logical thinking. 

Important Points to Remember while Studying JEE Main Mathematics

1. Dedicate more time to Algebra and Calculus. 
2. Attempt as many previous years’ question papers as you can for improved practice and time management skills. 
3. Keep your primary resource as the NCERT books. Use additional books for better practice. 

The points mentioned above will surely help you to ace through the topic in JEE Main 2020.