Probability of an event and basic probability rules

Probability is a fundamental concept in mathematics that deals with the likelihood of an event occurring. It quantifies uncertainty and is used extensively in various fields like science, engineering, finance, and even everyday decision-making. In the context of competitive exams like JEE Main, a solid understanding of probability is crucial.

What is an Event?

Before we dive into probability, let's define what an 'event' is. In probability theory, an event is a set of outcomes of an experiment to which a probability is assigned. An outcome is a single result of an experiment.

For example, consider the experiment of tossing a fair coin once.

• The sample space (the set of all possible outcomes) is S = {Head, Tail}.
• If we are interested in the event of getting a Head, we denote this event as A = {Head}.
• If we are interested in the event of getting a Tail, we denote this event as B = {Tail}.

Consider another experiment: rolling a fair six-sided die once.

• The sample space is S = {1, 2, 3, 4, 5, 6}.
• The event of getting an even number is E = {2, 4, 6}.
• The event of getting a number greater than 4 is G = {5, 6}.
• The event of getting a 3 is T = {3}.

What is Probability?

The probability of an event A, denoted by P(A), is a number between 0 and 1, inclusive, that measures how likely the event A is to occur.

• P(A) = 0 means the event A is impossible.
• P(A) = 1 means the event A is certain to occur.
• 0 < P(A) < 1 means the event A may or may not occur.

Types of Events

Understanding different types of events helps in solving probability problems more effectively.

1. Certain Event:

An event that is sure to happen. Its probability is 1. For example, when rolling a die, the event of getting a number less than 7 is a certain event, as all outcomes {1, 2, 3, 4, 5, 6} are less than 7.

2. Impossible Event:

An event that cannot happen. Its probability is 0. For example, when rolling a die, the event of getting a 7 is an impossible event.

3. Elementary Event:

An event that has only one outcome as favorable to it. For example, in rolling a die, the event of getting a '3' is an elementary event.

4. Compound Event:

An event that consists of two or more elementary events. For example, in rolling a die, the event of getting an even number {2, 4, 6} is a compound event.

5. Complementary Event:

If A is an event, then the event that A does not occur is called the complementary event of A, denoted by A' or A^c. The sum of the probabilities of an event and its complementary event is always 1.
P(A) + P(A') = 1
This implies P(A') = 1 - P(A).

6. Mutually Exclusive Events:

Two or more events are said to be mutually exclusive if the occurrence of one event precludes the occurrence of the other(s). In simpler terms, if one event happens, the others cannot happen at the same time. For example, when tossing a coin once, the events 'getting a Head' and 'getting a Tail' are mutually exclusive. You cannot get both a Head and a Tail simultaneously from a single toss.

7. Exhaustive Events:

A set of events is said to be exhaustive if at least one of them must occur. In other words, the union of exhaustive events covers the entire sample space. For example, in a single toss of a coin, the events 'getting a Head' and 'getting a Tail' are exhaustive because one of them is bound to happen.

Basic Probability Rules

There are several fundamental rules that govern probability calculations.

1. Classical Definition of Probability:

If an experiment has 'n' equally likely outcomes, and event A has 'm' of these outcomes as favorable to it, then the probability of event A is given by:
P(A) = (Number of favorable outcomes for A) / (Total number of possible outcomes)
P(A) = m / n

This definition is applicable only when the outcomes are equally likely.

Example: A bag contains 5 red balls and 3 blue balls. If a ball is drawn at random, what is the probability that it is red?

• Total number of balls = 5 (red) + 3 (blue) = 8.
• Number of favorable outcomes (drawing a red ball) = 5.
• Probability of drawing a red ball = P(Red) = 5 / 8.

2. Axiomatic Approach to Probability:

This approach defines probability based on a set of axioms (fundamental assumptions). Let S be the sample space of a random experiment. For any event A, the probability P(A) satisfies the following conditions:

1. P(A) ≥ 0 (Probability cannot be negative).
2. P(S) = 1 (The probability of the sample space, i.e., a certain event, is 1).
3. If A₁, A₂, A₃, ... are mutually exclusive events, then P(A₁ U A₂ U A₃ U ...) = P(A₁) + P(A₂) + P(A₃) + ...
   For two mutually exclusive events A and B, P(A U B) = P(A) + P(B).

3. Addition Rule of Probability:

This rule is used to find the probability of the union of two events. It states that for any two events A and B:
P(A U B) = P(A) + P(B) - P(A ∩ B)
Where P(A ∩ B) is the probability that both event A and event B occur.

Case 1: Mutually Exclusive Events
If A and B are mutually exclusive, then P(A ∩ B) = 0. The formula simplifies to:
P(A U B) = P(A) + P(B)
This is consistent with the third axiom.

Case 2: Non-Mutually Exclusive Events
If A and B can occur simultaneously, we must subtract the probability of their intersection.

Example: In a class of 30 students, 15 like Math, 10 like Science, and 5 like both Math and Science. What is the probability that a randomly selected student likes Math or Science?

• Let M be the event that a student likes Math. P(M) = 15/30.
• Let S be the event that a student likes Science. P(S) = 10/30.
• The event that a student likes both is M ∩ S. P(M ∩ S) = 5/30.
• We want to find P(M U S).
• Using the addition rule: P(M U S) = P(M) + P(S) - P(M ∩ S)
• P(M U S) = (15/30) + (10/30) - (5/30) = (15 + 10 - 5) / 30 = 20/30 = 2/3.

4. Multiplication Rule of Probability:

This rule is used to find the probability of the intersection of two events, P(A ∩ B). It is derived from the definition of conditional probability.

The rule states:
P(A ∩ B) = P(A) * P(B|A)
or
P(A ∩ B) = P(B) * P(A|B)
Here, P(B|A) is the conditional probability of event B occurring given that event A has already occurred. Similarly, P(A|B) is the conditional probability of event A occurring given that event B has already occurred.

Conditional Probability

Conditional probability is a crucial concept. P(B|A) is the probability of event B happening given that event A has already happened.

The formula for conditional probability is:
P(B|A) = P(A ∩ B) / P(A), provided P(A) > 0.

Similarly,
P(A|B) = P(A ∩ B) / P(B), provided P(B) > 0.

Example: A fair die is rolled. Let A be the event that the number rolled is even, and B be the event that the number rolled is greater than 3. Find P(B|A).

• Sample space S = {1, 2, 3, 4, 5, 6}.
• Event A (even number) = {2, 4, 6}. P(A) = 3/6 = 1/2.
• Event B (number > 3) = {4, 5, 6}. P(B) = 3/6 = 1/2.
• Event A ∩ B (even and > 3) = {4, 6}. P(A ∩ B) = 2/6 = 1/3.
• We want to find P(B|A) = P(A ∩ B) / P(A).
• P(B|A) = (1/3) / (1/2) = (1/3) * 2 = 2/3.

This means that if we know the number rolled is even, the probability that it is also greater than 3 is 2/3.

Independent Events

Two events A and B are said to be independent if the occurrence of one event does not affect the probability of the occurrence of the other event. Mathematically, events A and B are independent if:
P(A ∩ B) = P(A) * P(B)

If A and B are independent, then:
P(B|A) = P(B)
and
P(A|B) = P(A)

Example: Tossing a coin twice. Let A be the event of getting a Head on the first toss, and B be the event of getting a Head on the second toss.

• P(A) = 1/2.
• P(B) = 1/2.
• The outcome of the first toss does not affect the outcome of the second toss, so A and B are independent.
• P(A ∩ B) = P(getting Head on 1st AND Head on 2nd) = P(HH) = 1/4.
• Using the independence formula: P(A) * P(B) = (1/2) * (1/2) = 1/4.
• Since P(A ∩ B) = P(A) * P(B), the events are indeed independent.

It's important to distinguish between mutually exclusive events and independent events:

Probability of 'At Least One' and 'None'

These are common types of problems that can be efficiently solved using complementary probability.

1. Probability of 'At Least One':

The event 'at least one' means one or more occurrences. The complement of 'at least one' is 'none'.
P(at least one) = 1 - P(none)

Example: If a fair coin is tossed 3 times, what is the probability of getting at least one Head?

• Let H be the event of getting a Head, T be the event of getting a Tail. P(H) = 1/2, P(T) = 1/2.
• The experiment consists of 3 independent tosses.
• The complement of 'at least one Head' is 'no Heads', which means getting all Tails (TTT).
• P(all Tails) = P(T on 1st) * P(T on 2nd) * P(T on 3rd) = (1/2) * (1/2) * (1/2) = 1/8.
• P(at least one Head) = 1 - P(all Tails) = 1 - (1/8) = 7/8.

2. Probability of 'None':

This is straightforward. If you have a set of events, the probability of 'none' of them occurring is the probability that none of the desired outcomes happen.
P(none) = P(not A AND not B AND not C ...)

Example: From the previous example, the probability of getting no Heads is P(TTT) = 1/8.

Key Takeaways and Shortcuts

Practice Problems (Conceptual)

1. When rolling two dice, what is the probability that the sum of the numbers is 7?
(Hint: List all possible pairs that sum to 7. Total outcomes = 6 * 6 = 36.)
Favorable outcomes: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) - 6 outcomes.
P(Sum=7) = 6/36 = 1/6.

2. A card is drawn from a standard deck of 52 cards. What is the probability that it is a King or a Queen?
(Hint: These are mutually exclusive events.)
P(King) = 4/52. P(Queen) = 4/52.
P(King or Queen) = P(King) + P(Queen) = 4/52 + 4/52 = 8/52 = 2/13.

3. Three students, A, B, and C, are solving a problem independently. Their probabilities of solving it are 1/2, 1/3, and 1/4 respectively. What is the probability that the problem is solved?
(Hint: Use the complement rule. Find the probability that NONE of them solve it, then subtract from 1.)
P(A not solving) = 1 - 1/2 = 1/2.
P(B not solving) = 1 - 1/3 = 2/3.
P(C not solving) = 1 - 1/4 = 3/4.
P(None solve) = P(A not) * P(B not) * P(C not) = (1/2) * (2/3) * (3/4) = 6/24 = 1/4.
P(Problem is solved) = 1 - P(None solve) = 1 - 1/4 = 3/4.