Intermediate

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Probability — Basic Concepts, Addition & Multiplication Theorems

Master Probability Basic Concepts for JEE Main and Board Exams. Covers sample space events classical probability formula, complementary mutually exclusive exhaustive events, addition theorem P(A∪B)=P(A)+P(B)-P(A∩B), mutually exclusive special case, derived results for only A only B neither exactly one, multiplication theorem conditional probability P(B|A)=P(A∩B)/P(A), independent events P(A∩B)=P(A)P(B), at least one using complement method, odds in favour and against with verified practice questions.

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Probability is the mathematics of uncertainty — it provides a rigorous framework for quantifying how likely an event is to occur. From predicting the outcome of a dice roll to modelling real-world risk in science and engineering, probability is one of the most directly applicable branches of mathematics. In JEE and CBSE, probability problems test three core ideas: classical probability using equally likely outcomes, the Addition Theorem for the union of events, and the Multiplication Theorem for sequential or joint events. These tools, combined with a clear understanding of independent events, form the foundation on which all of probability theory is built.

1. Basic Terminology

Term	Definition	Example
Random Experiment	An experiment whose outcome cannot be predicted with certainty	Tossing a coin, rolling a die
Sample Space (S)	Set of all possible outcomes	S = {H, T} for coin; S = {1,2,3,4,5,6} for die
Event (A)	Any subset of the sample space	A = {even numbers} = {2,4,6}
Complementary Event (A')	All outcomes NOT in A	If A = {2,4,6} then A' = {1,3,5}
Mutually Exclusive	A and B cannot occur simultaneously: A ∩ B = ∅	Rolling a 3 and rolling a 5 on same die
Exhaustive Events	Events that together cover the entire sample space	A ∪ A' = S (always)

2. Classical (Theoretical) Probability

When all outcomes in the sample space are equally likely:

$P (A) = \frac{Number of favourable outcomes}{Total number of outcomes} = \frac{n (A)}{n (S)}$

Key properties:

0 ≤ P(A) ≤ 1 for any event A
P(S) = 1 (certain event); P(∅) = 0 (impossible event)
P(A') = 1 − P(A)

Standard Sample Spaces

Experiment	n(S)	Key counts
1 coin	2	1 head, 1 tail
2 coins	4	HH, HT, TH, TT; P(exactly 1H) = 2/4 = 1/2
1 die	6	P(even) = 3/6 = 1/2; P(>4) = 2/6 = 1/3
2 dice	36	P(sum=7) = 6/36 = 1/6; P(sum=11) = 2/36 = 1/18
Pack of 52 cards	52	13 each suit; 4 each rank; P(ace) = 4/52 = 1/13

3. Addition Theorem of Probability

$P (A \cup B) = P (A) + P (B) - P (A \cap B)$

Special case — Mutually Exclusive events (A ∩ B = ∅):

$P (A \cup B) = P (A) + P (B)$

Three events:

$P (A \cup B \cup C) = P (A) + P (B) + P (C) - P (A \cap B) - P (B \cap C) - P (A \cap C) + P (A \cap B \cap C)$

Useful Derived Results

Expression	Formula
P(only A, not B)	P(A) − P(A ∩ B)
P(only B, not A)	P(B) − P(A ∩ B)
P(exactly one of A, B)	P(A) + P(B) − 2P(A ∩ B)
P(neither A nor B)	1 − P(A ∪ B) = 1 − P(A) − P(B) + P(A ∩ B)
P(A ∩ B')	P(A) − P(A ∩ B)

Worked Example

P(A) = 3/10, P(B) = 2/5 = 4/10, P(A ∩ B) = 1/10. Find P(A ∪ B), P(only A), P(only B).

P(A ∪ B) = 3/10 + 4/10 − 1/10 = 6/10 = 3/5

P(only A) = 3/10 − 1/10 = 2/10 = 1/5

P(only B) = 4/10 − 1/10 = 3/10

4. Multiplication Theorem of Probability

For two events A and B:

$P (A \cap B) = P (A) \cdot P (B | A) = P (B) \cdot P (A | B)$

Where P(B|A) = conditional probability of B given A has occurred.

Conditional Probability (Basic Form)

$P (B | A) = \frac{P (A \cap B)}{P (A)}, P (A) \neq 0$

Independent Events

A and B are independent if the occurrence of one does not affect the other: P(B|A) = P(B). Then:

$P (A \cap B) = P (A) \cdot P (B)$

Note: Mutually exclusive ≠ Independent. In fact, if A ∩ B = ∅ and P(A), P(B) > 0, they are dependent (occurrence of one makes the other impossible).

"At Least One" Strategy

$P (at least one of A, B) = 1 - P (A^{'}) \cdot P (B^{'}) (for independent events)$

5. Odds For and Against an Event

If P(A) = p:

Odds in favour of A = p : (1−p) = P(A) : P(A')

Odds against A = (1−p) : p = P(A') : P(A)

If odds in favour = a : b, then P(A) = a/(a+b).

Probability Theorems — Quick Reference

Scenario	Formula
Any two events	P(A∪B) = P(A) + P(B) − P(A∩B)
Mutually exclusive	P(A∪B) = P(A) + P(B); P(A∩B) = 0
Complement	P(A') = 1 − P(A)
Conditional	P(B\|A) = P(A∩B)/P(A)
Multiplication	P(A∩B) = P(A)·P(B\|A)
Independent	P(A∩B) = P(A)·P(B)
At least one (independent)	1 − P(A')·P(B')
Odds a:b in favour	P(A) = a/(a+b)

JEE/Board Power Tips — Avoid These Common Mistakes

Mutually exclusive ≠ Independent. Mutually exclusive means P(A∩B) = 0 (they can't both happen). Independent means P(A∩B) = P(A)·P(B) (occurrence of one doesn't affect the other). Mutually exclusive events with nonzero probabilities are actually dependent.
P(A∪B) ≠ P(A) + P(B) unless A and B are mutually exclusive. Always check if P(A∩B) needs to be subtracted. Missing this is the most common error in probability problems.
For "at least one" problems, always use the complement: 1 − P(none). Direct calculation is error-prone; the complement method is faster and more reliable.
P(A∩B) = 0 does NOT mean A and B are independent. It means they are mutually exclusive (the opposite of independent in a probabilistic sense).
Cards: king of hearts is BOTH a king AND a heart. When applying the addition theorem P(king ∪ heart), subtract P(king of hearts) = 1/52. Many students forget this overlap.

Practice Questions

Q1 (JEE Main / Board): A card is drawn from a well-shuffled pack of 52 cards. Find the probability that it is either a king or a heart.

Explanation:

Let $A$ be the event of drawing a king, and $B$ be the event of drawing a heart.
$P (A) = \frac{4}{52} = \frac{1}{13}$
$P (B) = \frac{13}{52} = \frac{1}{4}$
$P (A \cap B)$ (drawing the king of hearts) $= \frac{1}{52}$

Using the addition theorem of probability:
$P (A \cup B) = P (A) + P (B) - P (A \cap B)$
$P (A \cup B) = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13}$

Q2 (Board): A bag contains 3 red and 2 blue balls. Two balls are drawn at random without replacement. Find the probability that both balls are red.

Explanation:

Method 1 (Combinatorics):
Total ways to draw 2 balls from 5 = $^{5} C_{2} = 10$
Ways to draw 2 red balls from 3 = $^{3} C_{2} = 3$
$P (both red) = \frac{3}{10}$

Method 2 (Conditional Probability):
$P (1st is red) = \frac{3}{5}$
$P (2nd is red | 1st was red) = \frac{2}{4}$
$P (both red) = (\frac{3}{5}) \times (\frac{2}{4}) = \frac{6}{20} = \frac{3}{10}$

Q3 (JEE Main): Given $P (A) = 0.6$ and $P (B) = 0.4$ , where $A$ and $B$ are independent events. Find the probability that at least one of the events $A$ or $B$ occurs.

Explanation:

Since $A$ and $B$ are independent, their complements $A^{'}$ and $B^{'}$ are also independent.
$P (A^{'}) = 1 - 0.6 = 0.4$
$P (B^{'}) = 1 - 0.4 = 0.6$

The probability that neither occurs is:
$P (neither) = P (A^{'} \cap B^{'}) = P (A^{'}) \times P (B^{'}) = 0.4 \times 0.6 = 0.24$

$P (at least one) = 1 - P (neither)$
$P (at least one) = 1 - 0.24 = 0.76$ (or $\frac{19}{25}$ )

Q4 (MCQ): If $P (A) = \frac{1}{2}$ , $P (B) = \frac{1}{3}$ , and $P (A \cap B) = \frac{1}{6}$ , are the events $A$ and $B$ independent?

A) Yes
B) No

Answer: A) Yes

Explanation: Two events are mathematically independent if and only if $P (A \cap B) = P (A) \times P (B)$ .
Checking the condition: $(\frac{1}{2}) \times (\frac{1}{3}) = \frac{1}{6}$ .
Since this perfectly matches the given $P (A \cap B)$ , the events are indeed independent.

Q5 (Board): Two fair dice are thrown simultaneously. Find the probability that the sum of the numbers appearing on the top faces is either 7 or 11.

Explanation:

Total number of outcomes in the sample space $n (S) = 6 \times 6 = 36$ .

Favorable pairs for a sum of 7: $(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) \to 6 outcomes$
Favorable pairs for a sum of 11: $(5, 6), (6, 5) \to 2 outcomes$

Since a sum cannot be both 7 and 11 simultaneously, these are mutually exclusive events.
Total favorable outcomes = $6 + 2 = 8$ .
$P (sum is 7 or 11) = \frac{8}{36} = \frac{2}{9}$

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Probability — Basic Concepts, Addition & Multiplication Theorems

1. Basic Terminology

2. Classical (Theoretical) Probability

Standard Sample Spaces

3. Addition Theorem of Probability

Useful Derived Results

Worked Example

4. Multiplication Theorem of Probability

Conditional Probability (Basic Form)

Independent Events

"At Least One" Strategy

5. Odds For and Against an Event

Probability Theorems — Quick Reference

Practice Questions

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