IIT JEE Main Maths -Unit 13-Basic probability rule- Study Notes-New Syllabus
IIT JEE Main Maths -Unit 13- Basic probability rule – Study Notes – New syllabus
IIT JEE Main Maths -Unit 13- Basic probability rule – Study Notes -IIT JEE Main Maths – per latest Syllabus.
Key Concepts:
- Basic probability rule
Probability
Probability measures how likely an event is to occur. It always lies between 0 and 1.
Classical Definition of Probability
\( P(E) = \dfrac{\text{Number of favorable outcomes}}{\text{Total number of equally likely outcomes}} \)
Used when all outcomes are equally likely (coins, dice, cards etc).
Basic Probability Rules
Rule 1: Probability of an Event
- \( 0 \le P(E) \le 1 \)
- \( P(E) = 1 \) → event always occurs
- \( P(E) = 0 \) → event never occurs
Rule 2: Complement of an Event
If \( E \) is an event, then complement \( E’ \) is the event that \( E \) does NOT happen.
\( P(E’) = 1 – P(E) \)
Rule 3: Addition Rule (Mutually Exclusive Events)
If events cannot happen together:
\( P(A \cup B) = P(A) + P(B) \)
Example: getting 2 or 5 on a dice.
Rule 4: Addition Rule (General Case)
\( P(A \cup B) = P(A) + P(B) – P(A \cap B) \)
Used when events may occur simultaneously.
Rule 5: Independent Events
If occurrence of one event does not affect the other:
\( P(A \cap B) = P(A)P(B) \)
Example: coin toss + dice roll.
Rule 6: Probability of Impossible and Certain Events
- Impossible event: \( P = 0 \)
- Certain event: \( P = 1 \)
Rule 7: Probability of At Least One Event
\( P(\text{at least one}) = 1 – (\text{probability that none occur}) \)
Very useful for JEE problems involving multiple trials.
Rule 8: Conditional Probability (Basic Form)
\( P(A|B) = \dfrac{P(A \cap B)}{P(B)} \)
Full conditional probability chapter will be covered later.
Example
A dice is rolled. Find the probability of getting an even number.
▶️ Answer / Explanation
Even numbers = 2, 4, 6 → 3 outcomes Total outcomes = 6
\( P(E) = \dfrac{3}{6} = \dfrac{1}{2} \)
Answer: \( \dfrac{1}{2} \)
Example
Two cards are drawn from a deck without replacement. Find the probability that both are aces.
▶️ Answer / Explanation
Number of aces = 4
\( P(\text{1st ace}) = \dfrac{4}{52} \)
After removing 1 ace: 3 aces left, 51 cards total \( P(\text{2nd ace}) = \dfrac{3}{51} \)
\( P(\text{both aces}) = \dfrac{4}{52} \cdot \dfrac{3}{51} = \dfrac{12}{2652} = \dfrac{1}{221} \)
Answer: \( \dfrac{1}{221} \)
Example
In a class, 60 percent students play cricket, 45 percent play football, and 20 percent play both. Find the probability that a randomly chosen student plays at least one of the two games.
▶️ Answer / Explanation
Use addition rule:
\( P(C \cup F) = P(C) + P(F) – P(C \cap F) \)
\( = 0.6 + 0.45 – 0.2 = 0.85 \)
Probability = 0.85
Notes and Study Materials
- Concepts of Probability
- Probability Master File
- Probability Notes
- Probability Formulae
- Probability Reference Book
- Probability Past Many Years Questions and Answer
Examples and Exercise
IIT JEE (Main) Mathematics ,”Probability” Notes ,Test Papers, Sample Papers, Past Years Papers , NCERT , S. L. Loney and Hall & Knight Solutions and Help from Ex- IITian
About this unit
Probability: Probability of an event, addition and multiplication theorems of probability, Baye’s theorem, probability distribution of a random variate, Bernoulli trials and Binomial distribution.
IITian Academy Notes for IIT JEE (Main) Mathematics – Probability
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IIT JEE (Main) Mathematics, Probability Solved Examples and Practice Papers.
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S. L. Loney IIT JEE (Main) Mathematics
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Hall & Knight IIT JEE (Main) Mathematics
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IIT JEE (Main) Mathematics Assignments
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Past Many Years (40 Years) Questions IIT JEE (Main) Mathematics Solutions Probability
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