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NCERT Class 11 Maths Chapter 14 Notes: Probability

NCERT Class 11 Maths Chapter 14, Probability, introduces the concept of chance and uncertainty in a mathematical way. It explains the basic terms like sample space, events, and the classical definition of probability. The chapter helps students understand how to calculate the probability of different events in a logical and systematic manner.

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 occur. 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
$A=\{\text { Ace of Hearts, Ace of Clubs, Ace of Diamonds, Ace of Spades }\}$

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

$\mathrm{E}_1 \cup \mathrm{E}_2 \cup \mathrm{E}_3 \cup \ldots \cup \mathrm{E}_n=\bigcup_{i=1}^n \mathrm{E}_i=\mathrm{S}$

then $\mathrm{E}_1, \mathrm{E}_2, \ldots, \mathrm{E}_n$ are called exhaustive events.
In other words, events $\mathrm{E}_1, \mathrm{E}_2, \ldots, \mathrm{E}_n$ of a sample space S are said to be exhaustive if at least one of them necessarily occurs 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 $\mathrm{E}_1, \mathrm{E}_2, \ldots, \mathrm{E}_n$ are $n$ events of a sample space S and if $\mathrm{E}_i \cap \mathrm{E}_j=\phi$ for every $i \neq j$, 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 $\mathrm{E}_1, \mathrm{E}_2, \ldots, \mathrm{E}_n$ 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 $A=\{1,4\}, \quad B=\{2,3,5\}, C=\{6\}$
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 $\mathrm{A} \cap \mathrm{B}=\mathrm{B} \cap \mathrm{C}=\mathrm{C} \cap \mathrm{A}=\phi$
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 $\phi$ and the sample space $S$ describe events. In fact, $\phi$ 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 $E_1=\{H H\}$ containing a single outcome $H H$ 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 $\bar{A}$. It is also called the event 'not $A$ '. Further $\mathrm{P}(\overline{\mathrm{A}})$ denotes the probability that A will not occur.

$\mathrm{A}^{\prime}=\overline{\mathrm{A}}=\mathrm{S}-\mathrm{A}=\{w: w \in \mathrm{~S}$ 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. $P: P(S) \rightarrow[0,1]$ satisfying the following axioms.
(i) For any event $\mathrm{E}, \mathrm{P}(\mathrm{E}) \geq 0$.
(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 $\mathrm{P}(\phi)=0$.
Let S be a sample space containing elementary outcomes $w_1, w_2, \ldots, w_n$, i.e., $\mathrm{S}=\left\{w_1, w_2, \ldots, w_n\right\}$

It follows from the axiomatic definition of probability that
(i) $0 \leq \mathrm{P}\left(w_i\right) \leq 1$ for each $w_i \in \mathrm{~S}$
(ii) $\mathrm{P}\left(w_i\right)+\mathrm{P}\left(w_2\right)+\ldots+\mathrm{P}\left(w_n\right)=1$
(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 favour 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 $\mathrm{S}=\left\{w_1, w_2, \ldots, w_n\right\}$ 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

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

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 $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{P}(\phi)=0$, 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 the NCERT Class 11 maths chapter 14 notes.

Probability: Previous Year Question and Answer

Given below are some previous year question answers of various examinations from the NCERT class 11 chapter 14, Probability:

Question 1: A die is loaded in such a way that each odd number is twice as likely to occur as each even number. Find P(G), where G is the event that a number greater than 3 occurs on a single roll of the die.

Solution:
Probability of odd nos. = 2 x (probability of even no.) ……….. (given)
Thus, P (Odd) = 2 x P (Even)
P(Odd) + P(Even) = 1
2P (Even) + P (Even) = 1
3P (Even) = 1
Thus, P (Even) = $\frac{1}{3}$
Thus, P (Odd) = 1 – $\frac{1}{3}$
$= \frac{3 – 1}{3}$
$= \frac{2}{3}$
Total no. occurring on a single roll = 6
& 4, 5 & 6 are the nos. greater than 3
Let P(no. greater than 3) = P (G)
= P (no. is 4, 5 or 6)
Here, 4 & 6 – Even & 5 – Odd
Thus, P (G) = 2 x P (Even) x P (Odd)
= 2 x 1/3 x 2/3
$=\frac{4}{9}$
Therefore, $\frac{4}{9}$ is the required probability.

Question 2: A card is drawn from a deck of 52 cards. Find the probability of getting a king or a heart or a red card.

Solution: We know that, in a deck,
Total no. of cards = 52
No. of kings = 4
No. of heart cards = 13
& total no. of red cards = 13 + 13 = 26
Thus, favorable outcomes = 4 + 13 + 26 – 13 – 2
= 28
Now,
Probability = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$
= $\frac{28}{52}$
= $\frac{7}{13}$

Question 3: In a non-leap year, the probability of having 53 Tuesdays or 53 Wednesdays is:

Solution: A non-leap year contains 365 days. So, on dividing it by 7, we get 52 weeks and 1 more day.
So, since 52 weeks are there, it means 52 Tuesdays will also be there, necessarily with probability I and 1 more day may be either Sun. or Mon. or Tue. or Wed. or Thur. or Fri. or Sat.
So, to get 53 Tuesdays, we have to select one more Tuesday from these 7 possibilities with a Probability of $\frac{1}{7}$.
Therefore, the probability of having 53 Tuesdays or 53 Wednesdays in a non-leap year $=\frac{1}{7}+ \frac{1}{7}=\frac{2}{7}$

Importance of NCERT Class 11 Maths Chapter 14 Notes

NCERT Class 11 Maths Chapter 16 notes will be very helpful for students to score good marks in their 11th-class exams. Students can download Probability Class 11 chapter 14 notes PDF to revise the concepts in a short period of time before the exam.

• 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 11 Maths Chapter 14 Probability.
• Students are advised to go through the NCERT Class 11 Maths Chapter 14 Notes before solving the questions.
• To boost your exam preparation as well as for quick revision, these NCERT notes are very useful.

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