Probability Class 11th Notes - Free NCERT Class 11 Maths Chapter 16 notes - Download PDF

Probability is a way to measure how likely something is to happen. For example, when you toss a coin, there’s an equal chance of getting heads or tails. So, the probability of getting heads is 1 out of 2. Probability is always a number between 0 and 1, both 0 and 1 inclusive. If something can’t happen, its probability is 0. If it will surely happen, its probability is 1. We use probability in real life to make guesses about things like the weather, winning a game, or making decisions. It helps us understand and handle situations where the result is not certain. Class 11 Maths Chapter 16 notes contain the following topics: sample space, outcomes, types of events, mutually exclusive events, exhaustive events, axiomatic approach, Addition rule, etc.

Class 11 Maths chapter 16 notes are prepared by subject matter experts in a precise manner so that students can get maximum benefit from them. The notes for Class 11 Maths Chapter 16 help a student to get a last-minute revision before the exam. Notes for Class 11 Maths Chapter 16 are made in such a way that no students face any difficulty during their preparation. NCERT Notes for Class 11 Maths cover the entire NCERT syllabus for this chapter.

NCERT CLASS 11 CHAPTER 16 NOTES

Random Experiment:

An experiment is random means that the experiment has more than one possible outcome, and it is not possible to predict with certainty which outcome will be. For instance, in an experiment of tossing an ordinary coin, it can be predicted with certainty that the coin will land either heads up or tails up, but it is not known for sure whether heads or tails will occur. If a die is thrown once, any of the six numbers, i.e., $1,2,3,4,5,6$ may turn up, not sure which number will come up.
(i) Outcome A possible result of a random experiment is called its outcome for example if the experiment consists of tossing a coin twice, some of the outcomes are $\mathrm{HH},\mathrm{HT}$, etc.
(ii) Sample Space A sample space is the set of all possible outcomes of an experiment. In fact, it is the universal set S pertinent to a given experiment.
The sample space for the experiment of tossing a coin twice is given by

$\mathrm{S}=\{\mathrm{HH},\mathrm{HT},\mathrm{TH},\mathrm{TT}\}$

The sample space for the experiment of drawing a card out of a deck is the set of all cards in the deck.

Outcomes:

The solution of the experiment is called the outcome.

Sample Space:

The set of all possible chances or outcomes is called the sample space.

Event:

An event is a subset of a sample space S. For example, the event of drawing an ace from a deck is

$A=\{\text{ Ace of Heart, Ace of Club, Ace of Diamond, Ace of Spade }\}$

Event ' $A$ or $B$ ': If A and B are two events associated with the same sample space, then the event 'A or B' is the same as the event $A\cup B$ and contains all those elements which are either in $A$ or in $B$ or both. Furthermore, $P(A\cup B)$ denotes the probability that A or B (or both) will occur.
Event ' $A$ and $B$ ': If $A$ and $B$ are two events associated with a sample space, then the event ' $A$ and $B$ ' is the same as the event $A\cap B$ and contains all those elements which are common to both A and B. Further more, $\mathrm{P}(\mathrm{A}\cap \mathrm{B})$ denotes the probability that both $A$ and $B$ will simultaneously occur.
The Event 'A but not $B$ ': (Difference $A-B$ ) An event $A-B$ is the set of all those elements of the same space $S$ which are in $A$ but not in $B$, i.e., $A-B=A\cap B^{\prime }$.
Mutually exclusive: Two events $A$ and $B$ of a sample space $S$ are mutually exclusive if the occurrence of any one of them excludes the occurrence of the other event. Hence, the two events $A$ and $B$ cannot occur simultaneously, and thus $P(A\cap B)=0$.

Exhaustive events: If ${\mathrm{E}}_1,{\mathrm{E}}_2,\dots ,{\mathrm{E}}_n$ are $n$ events of a sample space S and if

${\mathrm{E}}_1\cup {\mathrm{E}}_2\cup {\mathrm{E}}_3\cup \dots \cup {\mathrm{E}}_n=\bigcup _{i=1}^n{\mathrm{E}}_i=\mathrm{S}$

then ${\mathrm{E}}_1,{\mathrm{E}}_2,\dots ,{\mathrm{E}}_n$ are called exhaustive events.
In other words, events ${\mathrm{E}}_1,{\mathrm{E}}_2,\dots ,{\mathrm{E}}_n$ of a sample space S are said to be exhaustive if at least one of them necessarily occur whenever the experiment is performed.
Consider the example of rolling a die. We have $S=\{1,2,3,4,5,6\}$. Define the two events

A: 'a number less than or equal to 4 appears.'
B 'a number greater than or equal to 4 appears.'

Now
$A:\{1,2,3,4\},B=\{4,5,6\}$
$A\cup B=\{1,2,3,4,5,6\}=S$
Such events A and B are called exhaustive events.

Mutually exclusive and exhaustive events: If ${\mathrm{E}}_1,{\mathrm{E}}_2,\dots ,{\mathrm{E}}_n$ are $n$ events of a sample space S and if ${\mathrm{E}}_i\cap {\mathrm{E}}_j=\varphi$ for every $i\ne j$, i.e., ${\mathrm{E}}_i$ and ${\mathrm{E}}_j$ are pairwise disjoint and $\bigcup _{i=1}^n{\mathrm{E}}_i=\mathrm{S}$, then the events ${\mathrm{E}}_1,{\mathrm{E}}_2,\dots ,{\mathrm{E}}_n$ are called mutually exclusive and exhaustive events.

Consider the example of rolling a die.
We have $S=\{1,2,3,4,5,6\}$
Let us define the three events as
$A=a$ number which is a perfect square
$\mathrm{B}=$ a prime number
$\mathrm{C}=$ a number which is greater than or equal to 6
Now $A=\{1,4\},B=\{2,3,5\},C=\{6\}$
Note that $\mathrm{A}\cup \mathrm{B}\cup \mathrm{C}=\{1,2,3,4,5,6\}=\mathrm{S}$. Therefore, $\mathrm{A},\mathrm{B}$, and C are exhaustive events.
Also $\mathrm{A}\cap \mathrm{B}=\mathrm{B}\cap \mathrm{C}=\mathrm{C}\cap \mathrm{A}=\varphi$
Hence, the events are pairwise disjoint and thus mutually exclusive.
The classical approach is useful when all the outcomes of the experiment are equally likely. We can use logic to assign probabilities. To understand the classical method consider the experiment of tossing a fair coin. Here, there are two equally likely outcomes - head $(\mathrm{H})$ and tail $(\mathrm{T})$. When the elementary outcomes are taken as equally likely, we have a uniform probability model. If there are $k$ elementary outcomes in S, each is assigned the probability of $\frac{1}{k}$. Therefore, logic suggests that the probability of observing a head, denoted by $\mathrm{P}(\mathrm{H})$, is $\frac{1}{2}=0.5$, and that the probability of observing a tail, denoted $\mathrm{P}(\mathrm{T})$, is also $\frac{1}{2}=5$. Notice that each probability is between 0 and. Further, H and T are all the outcomes of the experiment and $\mathrm{P}(\mathrm{H})+\mathrm{P}(\mathrm{T})=1$.

Events are of different types:

(i) Impossible and Sure Events: The empty set $\varphi$ and the sample space $S$ describe events. In fact, $\varphi$ is called an impossible event and S, i.e., the whole sample space is called a sure event.

(ii) Simple or Elementary Event: If an event E has only one sample point of a sample space, i.e., a single outcome of an experiment, it is called a simple or elementary event. The sample space of the experiment of tossing two coins is given by

$\mathrm{S}=\{\mathrm{HH},\mathrm{HT},\mathrm{TH},\mathrm{TT}\}$

The event $E_1=\{HH\}$ containing a single outcome $HH$ of the sample space $S$ is simple. If one card is drawn from a well-shuffled deck, any particular card drawn like 'Queen of Hearts' is an elementary event.
(iii) Compound Event: If an event has more than one sample point it is called a compound event, for example, $\mathrm{S}=\{\mathrm{HH},\mathrm{HT}\}$ is a compound event.
(iv) Complementary event: Given an event $A$, the complement of $A$ is the event consisting of all sample space outcomes that do not correspond to the occurrence of $A$.

The complement of $A$ is denoted by $A$ ' or $\overline{A}$. It is also called the event 'not $A$ '. Further $\mathrm{P}(\bar{\mathrm{A}})$ denotes the probability that A will not occur.

${\mathrm{A}}^{\prime }=\bar{\mathrm{A}}=\mathrm{S}-\mathrm{A}=\{w:w\in \mathrm{S}$ and $w\notin \mathrm{A}\}$

Axiomatic Approach To Probability:

Let S be the sample space of a random experiment. The probability P is a real-valued function whose domain is the power set of $S$, i.e., $P(S)$ and range is the interval $[0,1]$ i.e. $P:P(S)\to [0,1]$ satisfying the following axioms.
(i) For any event $\mathrm{E},\mathrm{P}(\mathrm{E})\ge 0$.
(ii) $\mathrm{P}(\mathrm{S})=1$
(iii) If $E$ and $F$ are mutually exclusive events, then $P(E\cup F)=P(E)+P(F)$.

It follows from (iii) that $\mathrm{P}(\varphi )=0$.
Let S be a sample space containing elementary outcomes $w_1,w_2,\dots ,w_n$, i.e., $\mathrm{S}=\{w_1,w_2,\dots ,w_n\}$

It follows from the axiomatic definition of probability that
(i) $0\le \mathrm{P}\left(w_i\right)\le 1$ for each $w_i\in \mathrm{S}$
(ii) $\mathrm{P}\left(w_i\right)+\mathrm{P}\left(w_2\right)+\dots +\mathrm{P}\left(w_n\right)=1$
(iii) $\mathrm{P}(\mathrm{A})=\mathrm{P}\left(w_i\right)$ for any event A containing elementary events $w_i$

For example, if a fair coin is tossed once

$P(H)=P(T)=\frac{1}{2}$ satisfies the three axioms of probability.

Now suppose the coin is not fair and has double the chances of falling heads up as compared to the tails, then $\mathrm{P}(\mathrm{H})=\frac{2}{3}$ and $\mathrm{P}(\mathrm{T})=\frac{1}{3}$.
This assignment of probabilities is also valid for H and T as these satisfy the axiomatic definitions.

Probability of An Event:

S be the sample space, E be the event, and n(s)= n and n(E)=m

Then the probability of the event= P(E)

$P(E)=\frac{\text{ Number of favorable outcomes }}{\text{ total number of possible outcomes }}=\frac{m}{n}$

Odd number of outcomes in favor of the event: m : (n-m)

Odd number of outcomes against the event : (n-m): m

Probability of the event that does not occur or take place : P(A)= 1- P(A)

Probabilities of equally likely outcomes

Let a sample space of an experiment be $\mathrm{S}=\{w_1,w_2,\dots ,w_n\}$ and suppose that all the outcomes are equally likely to occur i.e., the chance of occurrence of each simple event must be the same

$\begin{array}{ll}\text{ i.e., }& \mathrm{P}\left(w_i\right)=p\text{ for all }{w}_i\in \mathrm{S},\text{ where }0\le p\le 1\\ \text{ Since }& {}^n\mathrm{P}\left(w_i\right)=1\\ \text{ i.e., }& p+p+p+\dots +p(n\text{ times })=1\\ & \Rightarrow np=1,\text{ i.e. }p=\frac{1}{n}\end{array}$

Let S be the sample space and E be an event, such that $n(\mathrm{S})=n$ and $n(\mathrm{E})=m$. If each outcome is equally likely, then it follows that

$\mathrm{P}(\mathrm{E})=\frac{m}{n}=\frac{\text{ Number of outcomes favourable to }\mathrm{E}}{\text{ Total number of possible outcomes }}$

Addition Rule :

Probability of event A or B :

P(A∪B) = P(A) + P(B) - P(A∩B)

Probability of event A or B or C :

P(A∪B∪C) = P(A) + P(B) +P(C)-P(A∩B)-P(B∩C) - P(A∩C) + P(A∩B∩C)

If A, B are mutually exclusive events: Then P(A∪B) = P(A) + P(B)

Since A∩B in an exclusive event is zero or null.

Addition rule for mutually exclusive events

If A and B are disjoint sets, then $\mathrm{P}(\mathrm{A}\cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})$ [since $\mathrm{P}(\mathrm{A}\cap \mathrm{B})=\mathrm{P}(\varphi )=0$, where A and B are disjoint]. The addition rule for mutually exclusive events can be extended to more than two events.

With this topic, we conclude NCERT Class 11 chapter 16 notes.

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