IIT JEE Main Maths -Unit 13- Addition and multiplication theorems- Study Notes-New Syllabus
IIT JEE Main Maths -Unit 13- Addition and multiplication theorems – Study Notes – New syllabus
IIT JEE Main Maths -Unit 13- Addition and multiplication theorems – Study Notes -IIT JEE Main Maths – per latest Syllabus.
Key Concepts:
- Addition and Multiplication Theorems of Probability
Addition and Multiplication Theorems of Probability
These two theorems are the foundation of probability. Nearly every JEE probability problem uses these formulas.
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Addition Theorem of Probability
(a) For Mutually Exclusive Events
If two events cannot occur together (like getting 2 or 5 on a die):
\( P(A \cup B) = P(A) + P(B) \)
Because
\( P(A \cap B) = 0 \)
(b) General Addition Rule (Events may occur together)
\( P(A \cup B) = P(A) + P(B) – P(A \cap B) \)
This is the most frequently used rule in JEE.
(c) Extension to Three Events
\( P(A \cup B \cup C) = P(A) + P(B) + P(C)\newline – P(A \cap B) – P(B \cap C) – P(C \cap A) + P(A \cap B \cap C) \)
Multiplication Theorem of Probability
The multiplication theorem connects intersection of events with independence or conditional probability.
(a) General Multiplication Theorem
For any two events:
\( P(A \cap B) = P(A)\cdot P(B|A) \)
And also:
\( P(A \cap B) = P(B)\cdot P(A|B) \)
(b) Independent Events
Two events are independent if occurrence of one does not affect the other.
\( P(A \cap B) = P(A)P(B) \)
This special case is used heavily in JEE.
(c) Extension to Three Independent Events
\( P(A \cap B \cap C) = P(A)P(B)P(C) \)
Valid only when all three are mutually independent.
Useful Identities
- \( P(A \cup B)’ = P(A’ \cap B’) \)
- \( P(A \cap B)’ = P(A’ \cup B’) \)
- \( P(A \cup B) = 1 – P(A’ \cap B’) \)
- \( P(\text{none}) = 1 – P(\text{at least one}) \)
Example
Two dice are rolled. Find the probability of getting at least one 6.
▶️ Answer / Explanation
Use addition theorem. Let:
\( A = \) first die shows 6
\( B = \) second die shows 6
\( P(A) = \dfrac{1}{6},\ P(B) = \dfrac{1}{6} \)
\( P(A \cap B) = \dfrac{1}{36} \)
\( P(A \cup B) = P(A) + P(B) – P(A \cap B) \)
\( = \dfrac{1}{6} + \dfrac{1}{6} – \dfrac{1}{36} = \dfrac{11}{36} \)
Answer: \( \dfrac{11}{36} \)
Example
A coin is tossed three times. Find the probability of getting exactly two heads.
▶️ Answer / Explanation
Independent events: each toss
Possible sequences with exactly 2 heads: HHT, HTH, THH → 3 outcomes
Total outcomes = 8
\( P = \dfrac{3}{8} \)
Answer: \( \dfrac{3}{8} \)
Example
In a class, 70% students pass in Maths, 60% in Physics, and 50% in both. Find the probability that a randomly chosen student passes in at least one subject.
▶️ Answer / Explanation
Use addition theorem:
\( P(M \cup P) = P(M) + P(P) – P(M \cap P) \)
\( = 0.7 + 0.6 – 0.5 = 0.8 \)
Answer: 0.8
