Probability Class 12th Notes - Free NCERT Class 12 Maths Chapter 13 Notes - Download PDF

Probability defines the chances of the occurrence of an event. For example chances of getting an even number when we roll a die are 50%, then its probability is 0.5. The probability of any event lies between 0 to 1, where 0 shows the absolute impossibility of an event, and 1 shows the maximum chances of an event certainly. Class 12 Maths chapter 13 notes contain the basics of probability, conditional probability, probability distribution, and all related theorems.

The main idea behind these Class 12 Maths Chapter 13 notes is to make the learning process easier for students and make a convenient revision tool whenever they need to recall important concepts and formulas. Experienced Careers360 experts prepared these notes, following the latest CBSE syllabus. Students should go through the NCERT textbook solutions first before checking these notes. For further practice, they can use NCERT exemplar solutions.

Probability:

In general terms, probability is defined as a measurement of the uncertainty of events in random experiments. Mathematically, it is the ratio of the number of outcomes to the total number of possible outcomes.

Conditional Probability

If E and F are two events associated with the same sample space of a random experiment, then the conditional probability of the event E under the condition that the event F has occurred, written as P(E∣F), is given by

P(E∣F)=P(E∩F)P(F),P(F)≠0

Properties of Conditional Probability

Property 1: P(F|F) = P(S|F) = 1

Property 2: If A and B are two events in the sample space S and F is an event of S such that P(F) ≠ 0, then

P((A ∪ B)|F) = P(A|F) + P(B|F) – P((A ∩ B)|F)

Property 3: P(E′|F) = 1 − P(E|F)

Multiplication Theorem on Probability

Let E and F be two events associated with an experiment sample space. Then

P(E∩F)=P(E)P(F∣E),P(E)≠0=P(F)P(E∣F),P(F)≠0

If E,F and G are three events associated with a sample space, then

P(E∩F∩G)=P(E)P(F∣E)P(G∣E∩F)

Independent Events

We have introduced conditional probability, where partial information that event B provides about event A. Now we found an interesting case where the occurrence of B does not provide information on A.

i.e. P(B|A)=P(B)

No w we can deduce that

P(B∣A)=P(B∩A)P(A),P(A)≠0P(B)=P(A∩B)P(A)P(A∩B)=P(A)⋅P(B)

Example:

In a dice. If M is the event in which a number appears as a multiple of 3, and N is the event in which even numbers occur. So, find whether events M and N are independent or not?

Solution:

We know that the sample space is S = {1, 2, 3, 4, 5, 6}

M={3,6}N={2,4,6}

There are even numbers in event M also.

M∩N={6}

Therefore,

P(M)=26=13P(N)=36=12P(M∩N)=13×12=16

So, P(M∩N)=P(M)⋅P(N)

From the above relation, we can say M and N are independent events.

Partition of a Sample Space

A set of events E1,E2,…,En is said to represent a partition of a sample space S if
(a) Ei∩Ej=ϕ,i≠j;i,j=1,2,3,……,n
(b) Ei∪E2∪…∪En=S, and
(c) Each Ei≠ϕ, i. e, P(Ei)>0 for all i=1,2,…,n

Theorem of Total Probability

Let {E1,E,…,En} be a partition of the sample space S . Let A be any event associated with S, then

P(A)=∑j=1nP(Ej)P(A∣Ej)

Bayes’ Theorem

Bayes' theorem is a theorem in probability that is used to determine the probability of the event that is related to an event that has already occurred.

The formula is

P(A∣B)=P(B∣A)P(A)P(B)

Probability Distribution of The Random Variable

Let S be a sample space associated with a random experiment. A function R: S→R is termed a random variable.

Example:

From 52 cards well-shuffled deck two cards are drawn. Find the probability for several aces.

Solution:

Let the number of aces be A.

Therefore, 0,1 or 2 will be the values of A.

When ace doesn’t occur

P(A=0)=P( non-ace )

P(A=0)=P( non − ace )×P( non − ace )=4852×4852=144169

When ace occurs once

P(A=1)=P( ace )⋅P( no )

P(A=1)=P( ace )⋅P( non − ace )+P( non − ace )⋅P( ace )=452×4852+4852×452=24169

When ace occurs twice

P(A=2)=P( ace − ace )=452×452=1169

The mean of a random variable

The mean of a random variable is used to locate the middle or the average value of the random variable.

μ=∑i=1nxiPi

The variance of a random variable

It is the expectation of the squared deviation of a random variable from its sample mean.

σ2=∑i=1n(xi−μ)2P(xi)

Bernoulli Trails and Bindraw Distribution

Bernoulli Trails

Trials of a random experiment are called Bernoulli trials, if they satisfy the following conditions:
(i) There should be a finite number of trials
(ii) The trials should be independent
(iii) Each trial has exactly two outcomes: success or failure
(iv) The probability of success (or failure) remains the same in each trial.

Example:

If there are 7 white and 9 red balls in a container, 6 balls are drawn successively. The trials of 6 balls from the container will be Bernoulli’s trials after each draw of the ball, as replaced or not replaced in the urn.

Solution:

1. If he trial numbers are finite then the drawing of the ball with replace, ent will be the success for drawing a white ball is ,p = 7/16. It will be the same for all six trials. So, the balls drawing with replacements are Bernoulli trials.

2. If the drawing of the ball is done without replacement, then the probability of success of drawing a white ball will be 7/16 for the first trial, 6/15 for the second trial, or if a red ball is drawn for 1st time will be 7/15, and so on. From the above statement, we can say that the success will not be the same for all the trials, so the trials are not Bernoulli trials.

Binomial Distribution

A random variable X taking values 0,1,2,…,n is said to have a binomial distribution with parameters n and p, if its probability distribution is given by

P(X=r)=ncrprqn−r,

where q=1−p and r=0,1,2,…,n.

Example:

Ten eggs are drawn successively with replacement from a lot containing 10% defective eggs. Find the probability that there is at least one defective egg.

Solution:

Let X denote the number of defective eggs in the 10 eggs drawn. Since the drawing is done with replacement, the trials are Bernoulli trials. X has the binomial distribution with

n=10 and 10%=10100,1 egg is defective out of 10 so 110

Therefore; q=1−p=010
Now P( at least 1 defect egg )=P(X⩾1)=1−P(X=0)

=1−10C0(910)10=1−9101010

This brings us to the end of the chapter

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Importance of Probability notes class 12

• NCERT problems are very important in order to perform well in exams. Students must try to solve all the NCERT problems, including miscellaneous exercises, and if needed, refer to the NCERT Solutions for Class 12 Maths Chapter 13 Probability.
• Students are advised to go through the NCERT Class 12 Maths Chapter 13 Notes before solving the questions.
• To boost your exam preparation as well as for quick revision, these NCERT notes are very useful.