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

NCERT Class 12 Maths Chapter 13 Notes: Free PDF Download

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NCERT Class 12 Maths Chapter 13 Notes: Probability

In general terms, the 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 experimental 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 have found an interesting case where the occurrence of B does not provide information on A.

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

Now 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 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 Trials and Binomial Distribution

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 the 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 Probability Class 12 Notes.

Probability: Previous Year Questions and Answers

Given below are some previous year question answers of various examinations from the NCERT class 12 chapter 13, Probability:

Question 1: Let a random variable X take values $0,1,2,3$ with $\mathrm{P}(\mathrm{X}=0)=\mathrm{P}(\mathrm{X}=1)=\mathrm{p}, \mathrm{P}(\mathrm{X}=2)=\mathrm{P}(\mathrm{X}=3)$ and $\mathrm{E}\left(\mathrm{X}^2\right)=2 \mathrm{E}(\mathrm{X})$. Then the value of $8 \mathrm{p}-1$ is :

Solution:
$\begin{aligned}
& P(X=0)=P(X=1)=p \text { and } P(X=2)=P(X=3)=q \\
& 2 p+2 q=1.......(1)\\
& \Rightarrow p+q=\frac{1}{2} \\
& E\left(X^2\right)=2 E(X) \\
& P\left(0^2\right)+p(1)^2+q(2)^2+q(3)^2 \\
& =2(p(0)+p(1)+q(2)+q(3)) \\
& \Rightarrow p=3q ........(2)
\end{aligned}$
From (1) and (2),
$\begin{aligned}
& q=\frac{1}{8} \text { and } p=\frac{3}{8} \\
&⇒ 8 p-1=2
\end{aligned}$
Hence, the correct answer is $2$.

Question 2: If $A$ and $B$ are two events such that $\mathrm{P}(\mathrm{A})=0.7$, $\mathrm{P}(\mathrm{B})=0.4$ and $\mathrm{P}(\mathrm{A} \cap \overline{\mathrm{B}})=0.5$, where $\overline{\mathrm{B}}$ denotes the complement of $B$, then $P(B \mid(A \cup \bar{B}))$ is equal:

Solution:

$\begin{aligned} & P(A)=0.7 \\ & P(B)=0.4 \\ & P\left(A \cap B^C\right)=0.5\end{aligned}$
$\begin{aligned}
& P\left(B / A \cup B^C\right)=\frac{P\left(B \cap\left(A \cup B^C\right)\right)}{P\left(A \cup B^C\right)} \\
& =\frac{P(A \cap B)}{P\left(A \cup B^C\right)}
\end{aligned}$
Using Venn diagram

$\Rightarrow \quad \frac{P(A \cap B)}{P\left(A \cup B^C\right)}=\frac{0.2}{0.5+0.2+0.1}=\frac{2}{8}=\frac{1}{4}$
Hence, the correct answer is $\frac{1}{4}$.

Question 3: The probability of forming a 12-person committee from 4 engineers, 2 doctors, and 10 professors containing at least 3 engineers and at least 1 doctor is:

Solution:
3 engineering + 1 doctor + 8 Professors $\rightarrow{ }^4 \mathrm{C}_3 \cdot{ }^2 \mathrm{C}_1 \cdot{ }^{10} \mathrm{C}_8$ $=360$
3 engineering +2 doctors +7 Professors $\rightarrow{ }^4 \mathrm{C}_3$. ${ }^2 \mathrm{C}_2 \cdot{ }^{10} \mathrm{C}_7$ $=480$
4 engineering +1 doctor +7 Professors $\rightarrow{ }^4 \mathrm{C}_4 \cdot{ }^2 \mathrm{C}_1 \cdot{ }^{10} \mathrm{C}_7$ $=240$
4 engineering +2 doctors +6 Professors $\rightarrow{ }^4 \mathrm{C}_4$. ${ }^2 \mathrm{C}_2 \cdot{ }^{10} \mathrm{C}_6$ $=210$
Total $=1290$
Required probability $=\frac{1290}{{ }^{16} \mathrm{C}_{12}}$
$=\frac{1290}{1820}$
$=\frac{129}{182}$
Hence, the correct answer is $\frac{129}{182}$.

NCERT Class 12 Notes Chapter Wise

All the links of chapter-wise notes for NCERT class 12 maths are given below

Subject Wise NCERT Exemplar Solutions

After finishing the textbook exercises, students can use the following links to check the NCERT exemplar solutions for a better clarity of the concepts.

• NCERT Exemplar Class 12 Solutions

• NCERT Exemplar Class 12 Maths

• NCERT Exemplar Class 12 Physics

• NCERT Exemplar Class 12 Chemistry

• NCERT Exemplar Class 12 Biology

Subject Wise NCERT Solutions

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• NCERT Solutions for Class 12 Mathematics

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• NCERT Solutions for Class 12 Physics

• NCERT Solutions for Class 12 Biology

NCERT Books and Syllabus

Students should always analyse the latest CBSE syllabus before making a study routine. The following links will help them check the syllabus. Also, give them access to more reference books.

• NCERT Book for Class 12
• NCERT Syllabus for Class 12