Probability is an important topic in Mathematics that deals with the likelihood of an event occurring. It helps you understand random experiments and calculate the chances of different outcomes.
For NIMCET preparation, you should understand sample spaces, events, probability formulas and different types of probability. Topics such as conditional probability, independent events and Bayes' theorem are also important to practise. You can also explore the NIMCET exam and its syllabus to understand the examination requirements.
Probability is a branch of Mathematics that measures the likelihood of an event occurring in a random experiment. Its value ranges from 0 to 1, where 0 represents an impossible event, and 1 represents a sure event. Understanding these basic concepts is important for your NIMCET exam preparation, as they form the foundation for solving different probability questions.
|
Particulars |
Details |
|
Meaning |
Measures the likelihood of an event occurring |
|
Probability Notation |
P(E) |
|
Minimum Value |
0 (Impossible Event) |
|
Maximum Value |
1 (Sure Event) |
|
General Probability Rule |
0 ≤ P(E) ≤ 1 |
|
Impossible Event |
An event that cannot occur |
|
Sure Event |
An event that always occurs |
Here, P(E) represents the probability of event E.
A random experiment is an activity that can produce different outcomes, but its exact outcome cannot be predicted with certainty.
A sample space (S) is the set of all possible outcomes of a random experiment. Each individual outcome is called a sample point.
|
Random Experiment |
Sample Space |
|
Tossing one coin |
{H, T} |
|
Rolling one die |
{1, 2, 3, 4, 5, 6} |
|
Tossing two coins |
{HH, HT, TH, TT} |
|
Rolling two dice |
36 ordered pairs |
An event is a subset of the sample space. It can contain one or more outcomes.
Simple Event: Contains exactly one sample point, such as getting 2 on a die.
Compound Event: Contains more than one sample point, such as getting an even number.
Exhaustive Events: A collection of events whose union covers the entire sample space.
Mutually Exclusive Events: Events that cannot occur at the same time.
For mutually exclusive events:
P(A ∩ B) = 0
For exhaustive events:
A₁ ∪ A₂ ∪ ... ∪ Aₙ = S
When all outcomes are equally likely, probability is calculated by dividing the number of favourable outcomes by the total number of possible outcomes.
Formula:
P(E) = Number of Favourable Outcomes / Total Number of Possible Outcomes
Solved Example: Find the probability of getting a multiple of 3 when rolling a die.
Favourable outcomes = {3, 6} = 2
Total possible outcomes = 6
Therefore,
P(E) = 2/6 = 1/3
Another example is selecting a vowel from the word PROBABILITY. It has 11 letters and 4 vowel positions (O, A, I, I).
P(Vowel) = 4/11
Repeated letters are counted separately when they occupy different positions.
Different probability formulas are used to calculate the chances of individual events, combined events and their complements.
The union of two events represents the occurrence of at least one event, while their intersection represents the occurrence of both events.
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
For mutually exclusive events:
P(A ∪ B) = P(A) + P(B)
The complement of an event represents the non-occurrence of that event.
P(A') = 1 − P(A)
These laws describe the relationship between the complements of unions and intersections.
(A ∪ B)' = A' ∩ B'
(A ∩ B)' = A' ∪ B'
The symmetric difference represents the occurrence of exactly one of two events.
P(A △ B) = P(A) + P(B) − 2P(A ∩ B)
Conditional Probability measures the probability of an event occurring when another event is already known to have occurred.
It is represented by P(A|B).
Formula:
P(A|B) = P(A ∩ B) / P(B), where P(B) > 0.
The multiplication theorem is derived from conditional probability.
P(A ∩ B) = P(A|B) × P(B)
Similarly,
P(A ∩ B) = P(B|A) × P(A)
Solved Example:
Given:
P(A) = 0.8
P(B) = 0.5
P(B|A) = 0.4
First, calculate the intersection:
P(A ∩ B) = 0.4 × 0.8 = 0.32
Now, calculate the conditional probability:
P(A|B) = 0.32/0.5 = 0.64
Finally, calculate the union:
P(A ∪ B) = 0.8 + 0.5 − 0.32 = 0.98
Two events are independent when the occurrence of one does not affect the probability of the other.
For independent events:
P(A ∩ B) = P(A) × P(B)
Other important relationships are:
P(A|B) = P(A)
P(B|A) = P(B)
Difference between Independent and Mutually Exclusive Events
Mutually exclusive events cannot occur together, whereas independent events do not affect each other's occurrence. Two events with positive probabilities cannot be both mutually exclusive and independent.
The Total Probability Theorem is used when an event can occur through multiple mutually exclusive and exhaustive cases.
If B₁, B₂, ..., Bₙ are mutually exclusive and exhaustive events, then:
P(A) = Σ P(A|Bᵢ)P(Bᵢ)
For three possible cases:
P(A) = P(A|B₁)P(B₁) + P(A|B₂)P(B₂) + P(A|B₃)P(B₃)
This theorem can be applied to problems involving multiple bags, different product sources and different match outcomes.
Bayes' Theorem is used to calculate reverse conditional probability. It helps identify a possible source or cause when an outcome is already known.
For mutually exclusive and exhaustive events B₁, B₂, ..., Bₙ:
P(Bᵢ|A) = [P(A|Bᵢ) × P(Bᵢ)] / [Σ P(A|Bⱼ) × P(Bⱼ)]
Important components include:
Prior Probability: P(Bᵢ), the initial probability of an event.
Observed Event Probability: P(A|Bᵢ), the probability of the observed event under a particular condition.
Posterior Probability: P(Bᵢ|A), the updated probability after observing the outcome.
Denominator: Total probability of the observed event.
Bayes' theorem is commonly applied to bag selection and ball-drawing problems, identifying the source of defective products and determining whether a person is telling the truth based on a reported outcome.
Questions based on playing cards and dice are useful for practising probability calculations.
A standard deck contains 52 cards divided into four suits, with 13 cards in each suit.
|
Card Category |
Number of Cards |
|
Hearts |
13 |
|
Diamonds |
13 |
|
Spades |
13 |
|
Clubs |
13 |
|
Red Cards |
26 |
|
Black Cards |
26 |
|
Kings |
4 |
|
Queens |
4 |
|
Jacks |
4 |
|
Face Cards |
12 |
|
Aces |
4 |
|
Honor Cards (Face Cards + Aces) |
16 |
Important Examples:
Probability of drawing a face card:
P(Face Card) = 12/52 = 3/13
Probability of drawing an honor card:
P(Honor Card) = 16/52 = 4/13
Probability of drawing a King or a Diamond:
P(K ∪ D) = P(K) + P(D) − P(K ∩ D)
= 4/52 + 13/52 − 1/52
= 4/13
The King of Diamonds belongs to both categories, so it is subtracted once to avoid double-counting.
Dice-based questions involve counting favourable outcomes from the possible results.
Solved Example: Find the probability of getting a sum greater than 7 when rolling two dice.
Favourable outcomes = 15
Total outcomes = 36
P(Sum > 7) = 15/36 = 5/12
Remember to consider ordered pairs when constructing the sample space for two dice.
The following table summarises the important probability formulas covered in this topic. Use it for quick revision and to recall the formulas while solving NIMCET Probability questions.
|
Concept |
Formula |
|
Probability Range |
0 ≤ P(A) ≤ 1 |
|
Classical Probability |
Favourable Outcomes / Total Outcomes |
|
Complement |
P(A') = 1 − P(A) |
|
Addition Theorem |
P(A ∪ B) = P(A) + P(B) − P(A ∩ B) |
|
Mutually Exclusive Events |
P(A ∩ B) = 0 |
|
Conditional Probability |
P(A|B) = P(A ∩ B)/P(B) |
|
Multiplication Theorem |
P(A ∩ B) = P(A|B)P(B) |
|
Independent Events |
P(A ∩ B) = P(A)P(B) |
|
Total Probability |
P(A) = Σ P(A|Bᵢ)P(Bᵢ) |
|
Bayes' Theorem |
P(Bᵢ|A) = P(A|Bᵢ)P(Bᵢ)/P(A) |
|
Symmetric Difference |
P(A △ B) = P(A) + P(B) − 2P(A ∩ B) |
While solving Probability Questions for NIMCET, it is important to understand the conditions of each formula and count outcomes carefully. Small calculation errors or confusion between probability concepts can affect the final answer. Keep the following common mistakes in mind while practising:
Counting the total possible outcomes incorrectly.
Confusing mutually exclusive events with independent events.
Applying conditional probability without checking whether the condition event has a non-zero probability.
Double-counting common outcomes while using the addition theorem.
Ignoring repeated letters or objects that represent different positions or outcomes.
Calculating the denominator incorrectly in Bayes' theorem.
Forgetting to consider ordered pairs when solving two-dice questions.
A clear understanding of basic concepts and regular question practice can help you prepare for Probability for NIMCET.
Follow these preparation points:
Understand sample spaces, events and event classifications before solving advanced questions.
Practise questions based on coins, dice, cards and balls.
Focus on conditional probability, independent events and the multiplication theorem.
Practise total probability and Bayes' theorem through bag-selection and source-identification problems.
Revise set operations and De Morgan's laws to solve event-based questions.
For organised preparation, explore NIMCET online batches and review the NIMCET exam pattern.
Probability for NIMCET includes basic concepts, event operations, conditional probability and important theorems. Focus on understanding the formulas and practising questions based on cards, dice and other random experiments. Regular revision of these concepts can help you strengthen your preparation for NIMCET Maths.
The classical probability formula is:
P(E) = Number of Favourable Outcomes / Total Number of Possible Outcomes.
