16.3 Event

Event

We have studied about random experiment and sample space associated with an experiment. The sample space serves as an universal set for all questions concerned with the experiment.
Consider the experiment of tossing a coin two times. An associated sample space is \(S =\{ HH , HT , TH , TT \}\).
Now suppose that we are interested in those outcomes which correspond to the occurrence of exactly one head. We find that \(HT \text { and } TH\) are the only elements of \(S\) corresponding to the occurrence of this happening (event). These two elements form the set \(E=\{ HT , TH \}\)
We know that the set \(E\) is a subset of the sample space \(S\). Similarly, we find the following correspondence Event between events and subsets of \(S\).
\(
\begin{array}{ll}
\text { Description of events } & \text { Corresponding subset of ‘S’ } \\
\text { Number of tails is exactly } 2 & A=\{ TT \} \\
\text { Number of tails is atleast one } & B =\{ HT , TH , TT \} \\
\text { Number of heads is atmost one } & C =\{ HT , TH , TT \} \\
\text { Second toss is not head } & D =\{ HT , TT \} \\
\text { Number of tails is atmost two } & S =\{ HH , HT , TH , TT \} \\
\text { Number of tails is more than two } & \phi
\end{array}
\)

Note: “At most two tails” means the event can have \(\mathbf{0}\) tails, or \(\mathbf{1}\) tail, or \(\mathbf{2}\) tails. Since the maximum number of tails when tossing two coins is 2 , all outcomes satisfy this condition.

The above discussion suggests that a subset of sample space is associated with an event and an event is associated with a subset of sample space. In the light of this we define an event as follows.

Definition: Any subset \(E\) of a sample space \(S\) is called an event.

Occurrence of an event

Consider the experiment of throwing a die. Let E denotes the event ” a number less than 4 appears”. If actually ‘ 1 ‘ had appeared on the die then we say that event \(E\) has occurred. As a matter of fact if outcomes are 2 or 3 , we say that event E has occurred

Thus, the event E of a sample space S is said to have occurred if the outcome \(\omega\) of the experiment is such that \(\omega \in E\). If the outcome \(\omega\) is such that \(\omega \notin E\), we say that the event E has not occurred.

Types of events

Events can be classified into various types on the basis of the elements they have.

Case-I: Impossible [Impossible Events (Probability = 0)] and Sure Events [Sure Events (Probability = 1)]

• The Dice Roll: Rolling a standard six-sided die and landing on a 7. Since the faces are only numbered 1 through 6, a 7 is physically impossible.
• The Dice Roll: Rolling a standard die and getting a number less than 7. Every possible outcome (1,2,3,4,5,6) satisfies this condition.

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 the sure event.

To understand these let us consider the experiment of rolling a die. The associated sample space is
\(
S=\{1,2,3,4,5,6\}
\)
Let \(E\) be the event ” the number appears on the die is a multiple of 7″. Can you write the subset associated with the event \(E\)?
Clearly no outcome satisfies the condition given in the event, i.e., no element of the sample space ensures the occurrence of the event E. Thus, we say that the empty set only correspond to the event E . In other words we can say that it is impossible to have a multiple of 7 on the upper face of the die. Thus, the event \(E =\phi\) is an impossible event.
Now let us take up another event F “the number turns up is odd or even”. Clearly
\(F =\{1,2,3,4,5,6\}=S\), i.e., all outcomes of the experiment ensure the occurrence of the event \(F\). Thus, the event \(F = S\) is a sure event.

Case-II: Simple Event

A Simple Event is an outcome that consists of exactly one single point in the sample space. In other words, it’s an event that cannot be broken down into simpler results.
Real-World Examples
Tossing a Coin:
When you flip a coin, the sample space is \(\{\) Heads, Tails \(\}\).
Simple Event: Landing specifically on Heads.
Simple Event: Landing specifically on Tails.
Rolling a Six-Sided Die:
The sample space is \(\{1,2,3,4,5,6\}\).
Simple Event: Rolling a 3.
Simple Event: Rolling a 6.
(Note: Rolling an “even number” is NOT simple because it contains three outcomes: 2, 4, and 6).

If an event \(E\) has only one sample point of a sample space, it is called a simple (or elementary) event.
In a sample space containing \(n\) distinct elements, there are exactly \(n\) simple events.
For example in the experiment of tossing two coins, a sample space is
\(
S =\{ HH , HT , TH , TT \}
\)
There are four simple events corresponding to this sample space. These are
\(
E _1=\{ HH \}, E _2=\{ HT \}, E _3=\{ TH \} \text { and } E _4=\{ TT \} \text {. }
\)

Case III: Compound Event

A Compound Event is an event that includes more than one outcome from the sample space. Essentially, it is a combination of two or more simple events.

Formula (Independent): \(P(A\) and \(B)=P(A) \times P(B)\).
Example: Tossing a coin (\(P=\frac{1}{2}\)) and rolling a 4 on a die (\(P=\frac{1}{6}\)) is a compound event with a probability of \(\frac{1}{2} \times \frac{1}{6}=\frac{1}{12}\).

Single Action Compound Events:
These happen when you perform one action (like one roll of a die) but look for a result that satisfies multiple possibilities.
Rolling a Die: Rolling an even number.
Outcomes: \(\{2,4,6\}\) (Three simple events combined).
Drawing a Card: Drawing a Face Card.
Outcomes: {Jack, Queen, King} of any suit (12 simple events combined).
Picking a Marble: In a bag of 10 marbles, picking a primary color (Red, Blue, or Yellow).

Multiple Action Compound Events:
These occur when you perform two or more actions in a row. The “event” is the specific combination of those results.
Two Coin Flips: Flipping a coin twice and getting at least one Head.
Outcomes: \(\{H T, T H, H H\}\)
A Snack Combo: Choosing a sandwich and a drink. The event of getting a “Turkey Sandwich AND Water” is compound because it relies on two separate outcomes happening together.
Rolling Two Dice: Rolling two dice and getting a sum of \(\mathbf{7}\).
Outcomes: \(\{(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\}\)

If an event has more than one sample point, it is called a Compound event.
For example, in the experiment of “tossing a coin thrice” the events
\(E\): ‘Exactly one head appeared’
\(F\): ‘Atleast one head appeared’
\(G\): ‘Atmost one head appeared’ etc.
are all compound events. The subsets of S associated with these events are
\(
\begin{aligned}
& E =\{\text { HTT,THT,TTH }\} \\
& F =\{\text { HTT,THT, TTH, HHT, HTH, THH, HHH }\} \\
& G =\{\text { TTT, THT, HTT, TTH }\}
\end{aligned}
\)
Each of the above subsets contain more than one sample point, hence they are all compound events.

Case-IV: Equally likely Events

Equally Likely Events are outcomes in a sample space that have the exact same theoretical probability of occurring. In simple terms, no single outcome has an “advantage” over the others; they all have a fair and equal shot. If an experiment has \(n\) possible outcomes and they are all equally likely, the probability of any one of those outcomes is:
\(
P(E)=\frac{1}{n}
\)

Definition: Two or more events are equally likely if they have the same probability of occurring.

Probability Calculation: For an event \({E}\) consisting of \({m}\) favorable outcomes within a sample space of \(n\) total equally likely, mutually exclusive, and exhaustive outcomes, the probability \(P(E)\) is \(\frac{m}{n}\).

Classic Examples
Tossing a Fair Coin:
In a single flip of a balanced coin, there are two outcomes: Heads and Tails.
Probability of Heads: 1/2 (50%)
Probability of Tails: \(1 / 2(50 \%)\)
Because the probabilities are identical, these are equally likely.

Rolling a Standard Die:
When you roll a fair six-sided die, there are six possible outcomes: \(\{1,2,3,4,5,6\}\).
Each number has a probability of exactly \(1 / 6\).
Since every number has the same chance, rolling a “2” is equally likely to rolling a “5.”

Case-V: Independent Events

Two events are Independent if the occurrence of one event does not change the probability of the other event happening. In simple terms, the two events have “no memory” of each other.
If Event A and Event B are independent, the mathematical rule is:
\(
P(A \text { and } B)=P(A) \times P(B)
\)

Definition: Two events \({A}\) and \({B}\) are independent if \({P}({A} \mid {B})={P}({A})\) or \(P(B \mid A)=P(B)\).
Multiplication Rule: \(P(A \cap B)=P(A) \times P(B)\).

Imagine you want to know the probability of flipping a coin and getting Heads, then rolling a die and getting a 6.
Probability of Heads: \(1 / 2\)
Probability of rolling a \(6: 1 / 6\)
Since they are independent:
\(
\frac{1}{2} \times \frac{1}{6}=\frac{1}{12}
\)

Case-VI: Complementary Event

A Complementary Event is the “opposite” of a specific event. If Event \(\boldsymbol{A}\) is what you are looking for, its complement (denoted as \(A^{\prime}\) or \(A^c\)) consists of all the other possible outcomes in the sample space that are not \(A\).

Formula: The probability of an event \(A\) not occurring is \(P\left(A^{\prime}\right)=1-P(A)\), where \(P\left(A^{\prime}\right)\) is the complement.

The most important rule to remember is that an event and its complement must add up to 1 (or \(100 \%\)), because together they cover every possibility.
Simple Examples:
The Coin Flip: If Event \(A\) is landing on Heads, the complement \(A^c\) is landing on Tails.
The Dice Roll: If Event \(A\) is rolling a 4 , the complement \(A^c\) is rolling a \(1,2,3,5\), or 6.
The Weather: If Event \(A\) is that it rains tomorrow, the complement \(A^c\) is that it does not rain tomorrow.

Let \(E\) be an event and \(S\) be the sample space for a random experiment, then complement of \(E\) is denoted by \(E^{\prime}\) or \(E^c\) or \(\bar{E}\). Clearly, \(E^{\prime}\) means \(E\) does not occur.
Thus, \(E^{\prime}\) occurs \(\Leftrightarrow E\) does not occur.
For example, When an unbiased die is thrown, then the sample space \(S=\{1,2,3,4,5,6\}\).
If
\(
E=\{1,4,6\} \text {, then } E^{\prime}=\{2,3,5\}
\)

Case-VII: Mutually Exclusive Events

Mutually Exclusive Events (also known as disjoint events) are events that cannot happen at the same time. If one event occurs, the other is automatically impossible for that specific trial.

Probability Formula: For two mutually exclusive events, the probability of either event occurring (A or B) is the sum of their individual probabilities:
\(
P(A \cup B)=P(A)+P(B) .
\)

Think of it as an “either-or” situation where there is zero overlap between the two possibilities.
Classic Examples
A Single Coin Flip: Landing on Heads and landing on Tails are mutually exclusive. You cannot get both on a single toss.
Rolling a Die: Rolling a 3 and rolling an even number are mutually exclusive. Since 3 is odd, it cannot also be even.
Turning a Corner: Turning Left and turning Right at the exact same intersection.
Deck of Cards: Drawing a card that is both a Heart and a Spade. Since every card belongs to only one suit, this is impossible.

A set of events is said to be mutually exclusive, if occurrence of one of them precludes the occurrence of any of the remaining events. If a set of events \(E_1, E_2, \ldots, E_n\) for mutually exclusive events.
Then,
\(
E_1 \cap E_2 \cap \ldots \cap E_n=\phi
\)

For example, If we thrown an unbiased die, then the sample space \(S=\{1,2,3,4,5,6\}\) in which \(E_1=\{1,2,3\}=\) the event of occurrence of a number less than 4 and \(E_2=\{5,6\}=\) the event of occurrence of a number greater than 4. Clearly, \(E_1 \cap E_2=\phi\). So, \(E_1\) and \(E_2\) are mutually exclusive.

Case-VIII: Exhaustive Events

Exhaustive Events are a set of events within a sample space where at least one of them must occur whenever the experiment is performed. In simpler terms, they “exhaust” all the possibilities.
If a set of events \(A, B\), and \(C\) are exhaustive, their combined probability must equal 1:
\(
P(A \cup B \cup C)=1
\)
Classic Examples
Rolling a Die: The events “Rolling an even number” \(\{2,4,6\}\) and “Rolling an odd number” \(\{1,3,5\}\) are exhaustive because every possible result (1 through 6 ) is covered.
The Weather: The events “It rains,” “It snows,” and “It is clear” are exhaustive if those are the only three defined states for that climate.
Exam Results: “Passing a test” and “Failing a test” are exhaustive events (assuming no other grade like “Incomplete” is possible).
Blood Types: The events having blood type A, B, AB, or O are exhaustive because every human fits into one of those categories.

A set of events is said to be exhaustive, if the performance of the experiment results in the occurrence of atleast one of them. If a set of events \(E_1, E_2, \ldots, E_n\) for exhaustive events.
Then, \(\quad E_1 \cup E_2 \cup \ldots \cup E_n=S\)
For example, If we thrown an unbiased die, then sample space \(S=\{1,2,3,4,5,6\}\) in which
\(E_1=\{1,2,3,4\}=\) the event of occurrence of a number less than 5 and \(E_2=\{3,4,5,6\}=\) the event of occurrence of a number greater than 2 .
Then, \(E_1 \cup E_2=\{1,2,3,4,5,6\}\) and \(E_1 \cap E_2=\{3,4\}\)
So, \(\quad E_1 \cup E_2=S\) 
Hence, \(E_1\) and \(E_2\) are exhaustive events.

Case-IX: Mutually Exclusive and Exhaustive Events

When events are both Mutually Exclusive and Exhaustive, they form a perfect partition of the sample space. This means every possible outcome belongs to exactly one of the events, and there is no overlap between them.
Mathematically, if events \(A\) and \(B\) are mutually exclusive and exhaustive:
No Overlap: \(P(A \cap B)=0\)
Total Coverage: \(P(A \cup B)=1\)
Real-World Examples:
The “Pass/Fail” System:
On a binary exam where you either pass or you don’t:
Mutually Exclusive: You cannot both pass and fail the same attempt.
Exhaustive: There is no third option; every student falls into one of these two categories.
Light Switches:
Consider a standard on/off light switch:
Mutually Exclusive: The switch cannot be “On” and “Off” at the same time.
Exhaustive: These are the only two states possible for the switch.
A Deck of Cards (by Color):
If you draw one card and define Event A as “Red Card” and Event B as “Black Card”:
Mutually Exclusive: A card cannot be both red and black.
Exhaustive: Every card in the deck is either red or black.

The Power of “1”:

When a set of events is both mutually exclusive and exhaustive, their probabilities must sum to exactly 1. This is the foundation of the Complement Rule.
\(
P(A)+P(B)+P(C) \ldots=1
\)
Example:
In a local election with three candidates (Smith, Jones, and Garcia), and assuming everyone must vote for exactly one person:
If \(P(\) Smith wins \()=0.3\)
And \(P(\) Jones wins \()=0.5\)
Then \(P\) (Garcia wins) must be 0.2 , because the events are mutually exclusive (only one winner) and exhaustive (someone must win).

A set of events is said to be mutually exclusive and exhaustive, if above two conditions are satisfied. If a set of events \(E_1, E_2, \ldots, E_n\) for mutually exclusive and exhaustive events.
Then, \(E_1 \cup E_2 \cup \ldots \cup E_n=S\) and \(E_1 \cap E_2 \cap \ldots \cap E_n=\phi\) For example, If we thrown an unbiased die, then sample space
\(S=\{1,2,3,4,5,6\}\) in which
\(E_1=\{1,3,5\}=\) the event of occurrence of an odd number and \(E_2=\{2,4,6\}=\) the event of occurrence of an even number.
Then, \(E_1 \cup E_2=\{1,2,3,4,5,6\}\) and \(E_1 \cap E_2=\phi\)
So, \(E_1 \cup E_2=S\) and \(E_1 \cap E_2=\phi\).
Hence, \(E_1\) and \(E_2\) are mutually exclusive and exhaustive events.

Algebra of events

In the Chapter on Sets, we have studied about different ways of combining two or more sets, viz, union, intersection, difference, complement of a set etc. Like-wise we can combine two or more events by using the analogous set notations.
Let \(A , B , C\) be events associated with an experiment whose sample space is \(S\).

Case-1: Complementary Event

For every event \(A\), there corresponds another event \(A ^{\prime}\) called the complementary event to \(A\). It is also called the event ‘not \(A\)‘.

For example, take the experiment ‘of tossing three coins’. An associated sample space is
\(
S =\{ HHH , HHT , HTH , THH , \text { HTT, THT, TTH, TTT }\}
\)
Let \(A =\{ HTH , HHT , THH \}\) be the event ‘only one tail appears’
Clearly for the outcome HTT, the event A has not occurred. But we may say that the event ‘not A’ has occurred. Thus, with every outcome which is not in A, we say that ‘not A’ occurs.
Thus the complementary event ‘not A ‘ to the event A is
\(
A ^{\prime}=\{ HHH , HTT , THT , TTH , TTT \}
\)
\(
\text { or } \quad A ^{\prime}=\{\omega: \omega \in S \text { and } \omega \notin A \}= S – A \text {. }
\)

Case-2: \(\text { The Event ‘A or B’ }\)

Recall that union of two sets \(A\) and \(B\) denoted by \(A \cup B\) contains all those elements which are either in \(A\) or in \(B\) or in both.
When the sets \(A\) and \(B\) are two events associated with a sample space, then ‘A \(\cup B\) ‘ is the event ‘either \(A\) or \(B\) or both’. This event ‘ \(A \cup B\) ‘ is also called ‘\(A\) or \(B\) ‘.
Therefore \(\quad\) Event ‘ A or B ‘= A \cup B[/latex]
\(=\{\omega: \omega \in A\) or \(\omega \in B\}\)

Case-3: \(\text { The Event ‘A and B’ }\)

We know that intersection of two sets \(A \cap B\) is the set of those elements which are common to both A and B. i.e., which belong to both ‘ \(A\) and \(B\) ‘.
If A and B are two events, then the set \(A \cap B\) denotes the event ‘ \(A\) and \(B\) ‘.
Thus, \(\quad A \cap B =\{\omega: \omega \in A\) and \(\omega \in B \}\)
For example, in the experiment of ‘throwing a die twice’ Let \(A\) be the event ‘score on the first throw is six’ and \(B\) is the event ‘sum of two scores is atleast 11 ‘ then \(A=\{(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}\), and \(B=\{(5,6),(6,5),(6,6)\}\)
so \(A \cap B =\{(6,5),(6,6)\}\)
Note that the set \(A \cap B =\{(6,5),(6,6)\}\) may represent the event ‘the score on the first throw is six and the sum of the scores is atleast 11’.

Case-4: \(\text { The Event ‘A but not B’ }\)

We know that \(A-B\) is the set of all those elements which are in \(A\) but not in \(B\). Therefore, the set \(A-B\) may denote the event ‘\(A\) but not \(B\) ‘. We know that
\(
A – B = A \cap B ^{\prime}
\)

Example 1: Consider the experiment of rolling a die. Let \(\text { A }\) be the event ‘getting a prime number’, \(\text { B }\) be the event ‘getting an odd number’. Write the sets representing the events (i) \(A\) or \(B\) (ii) \(A\) and \(B\) (iii) \(A\) but not \(B\) (iv) ‘not \(A\)‘.

Solution:

\(
\text { Here } \quad S =\{1,2,3,4,5,6\}, A =\{2,3,5\} \text { and } B =\{1,3,5\}
\)
Obviously
\(
\text { (i) ‘A or B’} = A \cup B =\{1,2,3,5\}
\)
\(
\text { (ii) ‘A and B’}= A \cap B =\{3,5\}
\)
\(
\text { (iii) ‘A but not B’} = A – B =\{2\}
\)
\(
\text { (iv) ‘not A’}= A ^{\prime}=\{1,4,6\}
\)

Case-5: Mutually exclusive events

In the experiment of rolling a die, a sample space is \(S =\{1,2,3,4,5,6\}\). Consider events, \(A\) ‘an odd number appears’ and \(B\) ‘an even number appears’

Clearly the event \(A\) excludes the event \(B\) and vice versa. In other words, there is no outcome which ensures the occurrence of events \(A\) and \(B\) simultaneously. Here
\(
A =\{1,3,5\} \text { and } B =\{2,4,6\}
\)

Clearly \(A \cap B =\phi\), i.e., \(A\) and \(B\) are disjoint sets.
In general, two events \(A\) and \(B\) are called mutually exclusive events if the occurrence of any one of them excludes the occurrence of the other event, i.e., if they can not occur simultaneously. In this case the sets \(A\) and \(B\) are disjoint.

Again in the experiment of rolling a die, consider the events \(A\) ‘an odd number appears’ and event \(B\) ‘a number less than 4 appears’
Obviously \(A =\{1,3,5\}\) and \(B =\{1,2,3\}\)

Now \(3 \in A\) as well as \(3 \in B\)
Therefore, \(A\) and \(B\) are not mutually exclusive events.

Remark: Simple events of a sample space are always mutually exclusive.

Case-6: Exhaustive events

Consider the experiment of throwing a die. We have \(S=\{1,2,3,4,5,6\}\). Let us define the following events
\(\text { A: }\) ‘a number less than 4 appears’,
\(\text { B: }\) ‘a number greater than 2 but less than 5 appears’ and
\(\text { C: }\) ‘a number greater than 4 appears’.
Then \(A =\{1,2,3\}, B =\{3,4\}\) and \(C =\{5,6\}\). We observe that \(A \cup B \cup C=\{1,2,3\} \cup\{3,4\} \cup\{5,6\}=S\).

Such events \(A , B\) and \(C\) are called exhaustive events. In general, if \(E _1, E _2, \ldots, E _{ n }\) are \(n\) events of a sample space \(S\) and if
\(
E _1 \cup E _2 \cup E _3 \cup \ldots \cup E _n=\cup_{i=1}^n E _i= S
\)
then \(E _1, E _2, \ldots ., E _{ n }\) are called exhaustive events. In other words, events \(E _1, E _2, \ldots, E _n\) are said to be exhaustive if atleast one of them necessarily occurs whenever the experiment is performed.
Further, if \(E _i \cap E _j=\phi\) for \(i \neq j\) i.e., events \(E _i\) and \(E _j\) are pairwise disjoint and \(\cup_{i=1}^n E _i= S\), then events \(E _1, E _2, \ldots, E _n\) are called mutually exclusive and exhaustive events.

Example 2: Two dice are thrown and the sum of the numbers which come up on the dice is noted. Let us consider the following events associated with this experiment
\(A\): ‘the sum is even’.
\(B\): ‘the sum is a multiple of 3 ‘.
\(C\): ‘the sum is less than 4 ‘.
\(D\): ‘the sum is greater than 11 ‘.
Which pairs of these events are mutually exclusive?

Solution: There are 36 elements in the sample space \(S =\{(x, y)\) : \(x, y=1,2,3,4,5,6\}\).
Then
\(
\begin{aligned}
A = & \{(1,1),(1,3),(1,5),(2,2),(2,4),(2,6),(3,1),(3,3),(3,5),(4,2),(4,4), \\
& (4,6),(5,1),(5,3),(5,5),(6,2),(6,4),(6,6)\} \\
B = & \{(1,2),(2,1),(1,5),(5,1),(3,3),(2,4),(4,2),(3,6),(6,3),(4,5),(5,4), \\
& (6,6)\} \\
C = & \{(1,1),(2,1),(1,2)\} \text { and } D =\{(6,6)\}
\end{aligned}
\)
We find that
\(
A \cap B=\{(1,5),(2,4),(3,3),(4,2),(5,1),(6,6)\} \neq \phi
\)

Therefore, \(A\) and \(B\) are not mutually exclusive events.
Similarly \(A \cap C \neq \phi, A \cap D \neq \phi, B \cap C \neq \phi\) and \(B \cap D \neq \phi\).
Thus, the pairs of events, \((A, C), (A, D), (B, C), (B, D)\) are not mutually exclusive events.
Also \(C \cap D =\phi\) and so \(C\) and \(D\) are mutually exclusive events.

Example 3: \(A\) coin is tossed three times, consider the following events.
\(A\): ‘No head appears’, \(B\): ‘Exactly one head appears’ and \(C\): ‘Atleast two heads appear’.
Do they form a set of mutually exclusive and exhaustive events?

Solution: The sample space of the experiment is
\(
S =\{ HHH , HHT , HTH , THH , HTT , THT , TTH , TTT \}
\)
\(
\text { and } A =\{ TTT \}, B =\{ HTT , THT , TTH \}, C =\{ HHT , HTH , THH , HHH \}
\)
Now \(A \cup B \cup C =\{ TTT , HTT , THT , TTH , HHT , HTH , THH , HHH \}= S\)
Therefore, \(A , B \text { and } C\) are exhaustive events. Also, \(A \cap B =\phi, A \cap C =\phi\) and \(B \cap C =\phi\)
Therefore, the events are pair-wise disjoint, i.e., they are mutually exclusive. Hence, \(A , B \text { and } C\) form a set of mutually exclusive and exhaustive events.

Favourable Elementary Events

Let \(S\) be the sample space associated with a random experiment and \(A\) be an event associated to the experiment. Then elementary events belonging to \(A\) are known as favourable elementary events to the event \(A\).
In order words, an elementary event \(E\) is said to be favourable to an event \(A\) if the occurrence of \(E\) ensures the happening or occurrence of event \(A\).
For example, in tossing of a die, 3 and 6 are elementary events favourable to the event “getting a multiple of 3 “.

Similarly, if a pair of dice is thrown, then \((2,6),(6,2),(4,4)\), \((3,5)\) and \((5,3)\) are elementary events favourable to the event “getting 8 as the sum”.
Consider the random experiment of drawing 4 cards from a well-shuffled pack of 52 playing cards. The total number of elementary events associated to this experiment is \({ }^{52} \mathrm{C}_4\), since 4 cards can be drawn out of 52 cards in \({ }^{52} \mathrm{C}_4\) ways. If we define the event \(A=\) Getting all red cards. Then, there are \({ }^{26} C_4\) elementary events favourable to this event, because 4 cards can be drawn out of 26 red cards in \({ }^{26} C_4\) ways. So, \({ }^{26} C_4\) is the fovourable number of elementary events to the event “getting 4 cards of the same colour”.

Probability

Definition: If there are \(n\) elementary events associated with a random experiment and \(m\) of them are favounable to an event \(A\), then the probability of happening or occurrence of \(A\) is denoted by \(P(A)\) and is defined as the ratio \(\frac{m}{n}\)
Thus, \(P(A)=\frac{m}{n}\)
Clearly, \(0 \leq m \leq n\)
\(
\therefore \quad 0 \leq \frac{m}{n} \leq 1 \Rightarrow 0 \leq P(A) \leq 1
\)
If \(P(A)=1\), then \(A\) is called certain event and \(A\) is called an impossible event, if \(P(A)=0\).
The number of elementary events which will ensure the nonoccurrence of \(A\) i.e. which ensure the occurrence of \(\bar{A}\) is \((n-m)\). Therefore,
\(
\begin{aligned}
P(\bar{A}) & =\frac{n-m}{n} \\
\Rightarrow \quad P(\bar{A}) & =1-\frac{m}{n} \Rightarrow P(\bar{A})=1-P(A) \Rightarrow P(A)+P(\bar{A})=1
\end{aligned}
\)
The odds in favour of occurrence of the event \(A\) are defined by \(m:(n-m)\) i.e \(; P(A): P(\bar{A})\) and the odds against the occurrence of \(A\) are defined by \(n-m: m\) i.e; \(P(\bar{A}): P(A)\).

Odds vs. Probability

This is where many people get tripped up, but your formulas are spot on. While probability is a ratio of the part to the whole, odds are a ratio of one part to another part.
\(
\begin{array}{lll}
\text { Concept } & \text { Formula / Ratio } & \text { Description } \\
\text { Probability } P(A) & \frac{m}{n} & \text { Favorable outcomes divided by total outcomes. } \\
\text { Odds in Favor } & m:(n-m) & \text { Successes compared to failures }(P(A): P(\bar{A})) . \\
\text { Odds Against } & (n-m): m & \text { Failures compared to successes }(P(\bar{A}): P(A)) .
\end{array}
\)

Example: If you have a \(20 \%\) chance of winning a game:
\(P(A)=0.2(\) or \(1 / 5)\)
Odds in favor are \(1: 4\) (For every 1 win, there are 4 losses).
Odds against are 4 : 1

Example 4: Three numbers are chosen from 1 to 30. The probability that they are not consecutive, is [CEE (Delhi) 2001]

Solution: Step 1: Total Number of Elementary Events (\(n\))
We are choosing 3 numbers from a set of 30 . The total number of ways to do this is given by the combination formula \({ }^n C_r\) :
\(
n={ }^{30} C_3=\frac{30 \times 29 \times 28}{3 \times 2 \times 1}=4060
\)
Step 2: Number of Favorable Events for “Consecutive” (\(m\))
First, we find how many sets of three numbers are consecutive. These sets are:
\(
(1,2,3),(2,3,4),(3,4,5), \ldots,(28,29,30)
\)
The sequence starts at 1 and ends at 28 . Therefore, there are exactly 28 such sets.
\(
m=28
\)
Step 3: Probability of Picking Consecutive Numbers
Using the definition \(P(A)=\frac{m}{n}\) :
\(
P(\text { Consecutive })=\frac{28}{4060}
\)
Simplifying the fraction (dividing both numerator and denominator by 28):
\(
P(\text { Consecutive })=\frac{1}{145}
\)
Step 4: Probability of Picking NOT Consecutive Numbers
Using the complement rule \(P(\bar{A})=1-P(A)\) :
\(
P(\text { Not Consecutive })=1-\frac{1}{145}=\frac{145-1}{145}=\frac{144}{145}
\)

Example 5: A die is tossed twice. The probability of having a number greater than 4 on each toss is

Solution: To solve this, we can use the definition of probability you provided earlier \(\left(P(A)=\frac{m}{n}\right)\), or we can treat the two tosses as independent events.
Step 1: Total Number of Elementary Events (\(n\))
When a die is tossed once, there are 6 possible outcomes. When it is tossed twice, the total number of outcomes is:
\(
n=6 \times 6=36
\)
Step 2: Favorable Outcomes for One Toss
For a single toss, the numbers greater than 4 are 5 and 6.
Number of favorable outcomes per toss = 2
Step 3: Favorable Outcomes for Both Tosses (\(m\))
Since we need a number greater than 4 on each toss, we can list the successful pairs:
\((5,5)\), \((5,6)\), \((6,5)\), \((6,6)\)
So, \(m=4\).
Step 4: Calculation
Using the formula:
\(
P(A)=\frac{m}{n}=\frac{4}{36}
\)
Simplifying the fraction:
\(
P(A)=\frac{1}{9}
\)
Alternative Method (Product Rule)
Since the two tosses are independent, you can simply multiply the probability of the first event by the probability of the second:
Probability of \(>4\) on 1 st toss: \(P\left(A_1\right)=\frac{2}{6}=\frac{1}{3}\)
Probability of \(>4\) on 2nd toss: \(P\left(A_2\right)=\frac{2}{6}=\frac{1}{3}\)
Total Probability: \(\frac{1}{3} \times \frac{1}{3}=\frac{1}{9}\)
The probability of having a number greater than 4 on each toss is \(1 / 9\).

Example 6: The probability that out of 10 persons, all born in June, at least two have the same birth day is:

Solution: To solve this, we use the “at least” principle. Calculating the probability of “at least two” directly is difficult because it includes the cases where \(2,3,4, \ldots\), or all 10 people share a birthday.
Instead, it is much simpler to find the probability that no two people share a birthday (all birthdays are different) and subtract that from 1.
Step 1: Total Number of Elementary Events (\(n\))
There are 10 people, and each person was born in June. June has 30 days.
Person 1 has 30 options.
Person 2 has 30 options.
… and so on for all 10 people.
\(
n=30^{10}
\)
Step 2: Favorable Events for the Complement (\(m\))
The complement event (\(\bar{A}\)) is that all 10 people have different birthdays.
Person 1 can be born on any of the \(\mathbf{3 0}\) days.
Person 2 must be born on a different day (29 options).
Person 3 must be born on a different day from the first two ( 28 options).
Continuing this for 10 people, we get:
\(
m=30 \times 29 \times 28 \times 27 \times 26 \times 25 \times 24 \times 23 \times 22 \times 21
\)
In permutation notation, this is written as \({ }^{30} C_{10}\).
Step 3: Probability that all birthdays are different \(P(\bar{A})\)
Using the definition \(P(\bar{A})=\frac{m}{n}\) :
\(
P(\bar{A})=\frac{{ }^{30} C_{10}}{30^{10}}
\)
Step 4: Probability of at least two having the same birthday \(P(A)\) Using the rule \(P(A)=1-P(\bar{A})\) :
\(
P(A)=1-\frac{{ }^{30} C_{10}}{30^{10}}
\)

Example 7: Two number \(b\) and \(c\) are chosen at random with replacement from the numbers \(1,2,3,4,5,6,7,8\) and 9. The probability that \(x^2+b x+c>0\) for all \(x \in R\) is?

Solution: To solve this, we need to combine your definition of probability with the algebraic condition for a quadratic expression to always be positive.
Step 1: The Algebraic Condition
For a quadratic expression \(a x^2+b x+c\) with \(a>0\) to be greater than zero for all \(x \in R\), its discriminant ( \(D\) ) must be less than zero. This means the parabola sits entirely above the \(x\)-axis and never touches or crosses it.
\(
D=b^2-4 a c<0
\)
In our case, \(a=1\), so the condition is:
\(
b^2-4 c<0 \Longrightarrow b^2<4 c
\)
Step 2: Total Number of Elementary Events (\(n\))
Since \(b[latex] and [latex]c\) are chosen from the set \(\{1,2,3,4,5,6,7,8,9\}\) with replacement:
Possible values for \(b=9\)
Possible values for \(c=9\)
\(
n=9 \times 9=81
\)
Step 3: Finding Favorable Outcomes (\(m\))
We need to count all pairs \((b, c)\) such that \(b^2<4 c\). We can test each possible value of \(b\) :
If \(b=1: b^2=1\). We need \(1<4 c\). This is true for \(c \in\{1,2, \ldots, 9\}\). (9 cases)
If \(b=2: b^2=4\). We need \(4<4 c\), so \(c>1 . c \in\{2,3, \ldots, 9\}\). (8 cases)
If \(b=3: b^2=9\). We need \(9<4 c\), so \(c>2.25\). \(c \in\{3,4, \ldots, 9\}\). ( 7 cases)
If \(b=4: b^2=16\). We need \(16<4 c\), so \(c>4\). \(c \in\{5,6,7,8,9\}\).(5 cases)
If \(b=5: b^2=25\). We need \(25<4 c\), so \(c>6.25\). \(c \in\{7,8,9\}\). (3 cases)
If \(b=6: b^2=36\). We need \(36<4 c\), so \(c>9\). (0 cases) (Since the max value for \(c\) is 9 , no values for \(b \geq 6\) will work.)
Total favorable outcomes (\(m\)):
\(
m=9+8+7+5+3=32
\)
Step 4: Final Probability Calculation
Using your formula \(P(A)=\frac{m}{n}\) :
\(
P\left(x^2+b x+c>0\right)=\frac{32}{81}
\)

Example 8: A \(2 \times 2\) square matrix is written down at random using the numbers \(1,-1\) as elements. The probability that the matrix is non-singular is ?

Solution: To find the probability that a \(2 \times 2\) matrix is non-singular, we need to identify how many total matrices can be formed and how many of those have a non-zero determinant.
Step 1: Total Number of Elementary Events (\(n\))
A \(2 \times 2\) matrix has 4 positions. Each position can be filled by one of 2 numbers ( 1 or -1 ).
\(
n=2 \times 2 \times 2 \times 2=2^4=16
\)
Step 2: Condition for Non-Singular Matrix
A matrix \(A=\left(\begin{array}{ll}a & b \\ c & d\end{array}\right)\) is non-singular if its determinant is not zero:
\(
\operatorname{det}(A)=a d-b c \neq 0
\)
Given \(a, b, c, d \in\{1,-1\}\), the product \(a d\) can be 1 or -1 , and the product \(b c\) can also be 1 or -1.
The determinant \(a d-b c\) can only result in three possible values:
\(1-1=0\) (Singular)
\(-1-(-1)=0\) (Singular)
\(1-(-1)=2\) (Non-singular)
\(-1-1=-2\) (Non-singular)
Step 3: Finding Favorable Outcomes (\(m\))
It is often faster to count the singular cases (\(a d=b c\)) and subtract them from the total.
Cases where \(a d=b c\) :
\(a d=1\) and \(b c=1\) :
\(a d=1\) happens if \((a, d)\) is \((1,1)\) or \((-1,-1)\) [2 ways]
\(b c=1\) happens if \((b, c)\) is \((1,1)\) or \((-1,-1)\) [2 ways]
Total: \(2 \times 2=4\) ways.
\(a d=-1\) and \(b c=-1\) :
\(a d=-1\) happens if \((a, d)\) is \((1,-1)\) or \((-1,1)\) [2 ways]
\(b c=-1\) happens if \((b, c)\) is \((1,-1)\) or \((-1,1)\) [2 ways]
Total: \(2 \times 2=4\) ways.
Total singular matrices \(=4+4=8\).
Favorable (Non-singular) matrices (\(m\)):
\(
m=\text { Total }- \text { Singular }=16-8=8
\)
Step 4: Final Probability Calculation
Using the definition \(P(A)=\frac{m}{n}\) :
\(
P(\text { Non-singular })=\frac{8}{16}=\frac{1}{2}
\)

Alternate: A \(2 \times 2\) square matrix has 4 elements each of which can be chosen in 2 ways.
∴ Total number of \(2 \times 2\) square matrices with elements 1 and – 1
\(
=2^4=16 .
\)
Out of these 16 matrices, following matrices are singular:
\(
\begin{aligned}
& {\left[\begin{array}{rr}
1 & 1 \\
-1 & -1
\end{array}\right],\left[\begin{array}{rr}
-1 & -1 \\
1 & 1
\end{array}\right],\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right],\left[\begin{array}{ll}
-1 & -1 \\
-1 & -1
\end{array}\right],} \\
& {\left[\begin{array}{ll}
-1 & 1 \\
-1 & 1
\end{array}\right],\left[\begin{array}{ll}
1 & -1 \\
1 & -1
\end{array}\right],\left[\begin{array}{rr}
1 & -1 \\
-1 & 1
\end{array}\right],\left[\begin{array}{rr}
-1 & 1 \\
1 & -1
\end{array}\right],}
\end{aligned}
\)
∴ \(\quad\) Number of non-singular matrices \(=16-8=8\).
Hence, required probability \(=\frac{8}{16}=\frac{1}{2}\)

Example 9: Two numbers \(a\) and \(b\) are selected at random from \(1,2, \ldots ., 100\) and are multiplied. The probability that the product \(a b\) is divisible by 3 , is

Solution: To find the probability that the product \(a b\) is divisible by 3 , it is easiest to use the Complement
Rule: \(P\) (divisible by 3 ) \(=1-P\) (not divisible by 3).
A product \(a b\) is not divisible by 3 if and only if neither \(a\) nor \(b\) is divisible by 3.
Step 1: Categorize the Numbers
In the set \(\{1,2, \ldots, 100\}\) :
Numbers divisible by 3:3,6,9, . . , 99.
To find the count: \(99=3 k \Longrightarrow k=33\).
There are \(\mathbf{3 3}\) such numbers.
Numbers NOT divisible by 3: \(100-33=67\).
There are 67 such numbers.
Step 2: Total Number of Elementary Events (\(n\))
We are selecting 2 numbers from 100 . Assuming the selection is made without replacement (common for “selecting numbers at random” unless “with replacement” is specified):
\(
n={ }^{100} C_2=\frac{100 \times 99}{2}=4950
\)
Step 3: Favorable Outcomes for the Complement (\(m\))
For the product \(a b\) to not be divisible by 3 , both \(a\) and \(b\) must be chosen from the 67 numbers that are not divisible by 3.
\(
m={ }^{67} C_2=\frac{67 \times 66}{2}=67 \times 33=2211
\)
Step 4: Final Probability Calculation
Probability that the product is NOT divisible by 3 :
\(
P(\bar{A})=\frac{2211}{4950}
\)
Divide both by \(33: \frac{67}{150}\).
Probability that the product IS divisible by 3:
Using \(P(A)=1-P(\bar{A})\) :
\(
\begin{gathered}
P(A)=1-\frac{67}{150} \\
P(A)=\frac{150-67}{150}=\frac{83}{150}
\end{gathered}
\)

Alternate: Two numbers can be chosen out of first 100 natural numbers in \({ }^{100} C_2\) ways.
\(\therefore \quad\) Total number of elementary events \(={ }^{100} C_2\)
The product \(a b\) is divisible by 3 if at least one of the two numbers is divisible by 3.
\(\therefore \quad\) Required probability
\(=1\) – Probability that none of the two numbers chosen is divisible by 3.
\(
=1-\frac{{ }^{67} C_2}{{ }^{100} C_2}=1-\frac{67 \times 33}{50 \times 99}=1-\frac{67}{150}=\frac{83}{150}
\)

Example 10: Each coefficient in the equation \(a x^2+b x+c=0\) is determined by throwing an ordinary six faced die. The probability that the equation will have real roots, is?

Solution: To find the probability that the quadratic equation \(a x^2+b x+c=0\) has real roots, we look at the coefficients \(a, b\), and \(c\), which are each determined by a die roll (values 1 through 6).
Step 1: The Condition for Real Roots
For a quadratic equation to have real roots, the discriminant (\(D\)) must be greater than or equal to zero:
\(
D=b^2-4 a c \geq 0 \Longrightarrow b^2 \geq 4 a c
\)
Step 2: Total Number of Elementary Events (\(n\))
Since each of the three coefficients \((a, b, c)\) is determined by an independent throw of a sixsided die:
\(
n=6 \times 6 \times 6=216
\)
Step 3: Finding Favorable Outcomes (\(m\))
We need to count all combinations \((a, b, c)\) where \(b^2 \geq 4 a c\). We will iterate through possible values of \(b\) and count valid pairs of (\(a, c\)).

Total favorable outcomes (\(m\)):
\(
m=0+1+3+8+14+17=43
\)
Step 4: Final Probability Calculation
Using the formula \(P(A)=\frac{m}{n}\) :
\(
P(\text { Real Roots })=\frac{43}{216}
\)
The probability that the equation has real roots is \(\frac{43}{216}\).