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

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

Edited By Ramraj Saini | Updated on Apr 23, 2022 01:32 PM IST

Probability Class 12 notes belong to the 13 chapter of NCERT. The NCERT Class 12 Maths chapter 13 notes are entirely based on the important topics required for the exam. Class 12 Maths chapter 13 notes nicely define the important formulas and their required derivations. The notes for Class 12 Maths chapter 13 helps a student to get a last-minute revision before the exam. Notes for Class 12 Maths chapter 13 is made in such a way that no students face any difficulty during their preparation. NCERT Notes for Class 12 Maths chapter 13 not only cover the NCERT notes but also covers the CBSE Class 12 Maths chapter 13 notes.

After going through Class 12 Probability notes

students can also refer to,

NCERT Class 12 Maths Chapter 13 Notes

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

It gives us the way to find a reason for the experiment based on partial information. Let us think of such a situation that like

  1. In a consecutive roll of dice, the sum is 9. What is the probability that the first roll is 6 and many more?

Thus, we can define the conditional probability that if A and B are two events of sample space S and P(A)≠0, then the probability of B after the event A has occurred is called conditional Probability.

\\ Formula:\\ P(B|A)=\frac{P(B\cap A)}{P(A)}, \ \ \ P(A) \neq 0 \\ \\ P(B|A)= \frac{n(A \cap B)}{n(A)}

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 being 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

For Given two dependent events A and B

P(A ∩ B) = P(A) P(B | A)

The multiplication rules of probability for more than two given events are as follows

Let E, F and G are three events given in a sample space, then we have

P(E ∩ F ∩ G) = P(E) P(F|E) P(G|(E ∩ F)) = P(E) P(F|E) P(G|EF)

Similarly, this rule can be extended for four or more events.

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)

Now we can deduce that

\\ P(B|A)=\frac{P(B\cap A)}{P(A)}, \ \ \ P(A) \neq 0 \\ \\ P(B)= \frac{P(A \cap B)}{P(A)} \\ \\ P(A\cap B)=P(A).P(B)

Example:

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

Solution:

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

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There are even numbers in event M also.

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Therefore,

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So, 1646636992877

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

Bayes’ Theorem

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

The formula is

P(A|B)= \frac{P(B|A)P(A)}{P(B)}

Total Probability Theorem

\\ Let \ (E_1, E_2,..., E_n )\ \text{ be a portion of the sample space S, and let us assume that each of the even} \\ E_1, E_2,..., E_n \text{ has nonzero occurrence probability. Then let us consider A to be an event associated with S, then} \\ P(A) = P(E_1 ) P(A|E_1 ) + P(E_2 ) P(A|E_2 ) + ... + P(E_n ) P(A|E_n )\\ \\ = \sum_{j=1}^{n}P(E_j)P(A|E_j)

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 a number of 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

1646639349439

1646639349921

When ace occurs once

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When ace occurs twice

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X

0

1

2

P(X)

144/169

24/169

1/169


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.

\mu = \sum_{i=1}^{n}x_iP_i

The variance of a random variable

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

\sigma^2=\sum_{i=1}^{n}(x_i-\mu)^2P(x_i)

Bernoulli Trails and Binomial Distribution

Bernoulli Trails

Trials that contain only two outcomes usually referred to as ‘success’ or ‘failure’ are known as Bernoulli trials.

Example:

If there are 7 white and 9 red balls in a container 6 balls are drawn successively. Find out the trails of 6 balls from the container will be Bernoulli’s trials after each draws 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 replacement 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 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

It is a type of distribution that entire up the likely values that will take one or two independent values given under a set of assumptions.

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. Clearly, X has the binomial distribution with

\\ n = 10\ and\ 10% =\frac{10}{100} \ , \text{1 egg is defective out of 10 so} \ \frac{1}{10} \\ \\ Therefore;\ q=1-p = \frac{0}{10} \\ \\ Now \ P( \text{at least 1 defect egg} ) = P(X\geqslant 1) =1-P(X=0) \\ \\ =1 - \ ^{10}C_{0} \ (\frac{9}{10})^{10} =1-\frac{9^{10}}{10^{10}}

This brings us to the end of the chapter

Significance of NCERT Class 12 Math Chapter 13 Notes

Class 12 Probability notes will be really helpful for the revision of the entire chapter and getting a list of the important topics covered in the notes. Also, Class 12 Math chapter 13 Notes is useful for getting a glance of Class 12 CBSE Syllabuses and also for national competitive exams like BITSAT, and JEE MAINS. Class 12 Maths chapter 13 notes pdf download can be used for getting a hard copy and to prepare for the exam.

NCERT Class 12 Notes Chapter Wise.

Frequently Asked Questions (FAQs)

1. Define Probability according to Probability Class 12 notes.

It is a department of maths that deals with the occurrence of random event

2. What is probability’s formula

It is defined as the possibility of an event to happen is equal to the ratio of the number of outcomes and the total number of outcomes.

3. Name the textbook that to be followed

Class 12 Math chapter 13 textbook should be followed.

4. From where we can download these notes

The notes can be downloaded from probability Class 12 notes pdf download.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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