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JEE Advanced Geometry: Circles — practice questions

Explore 2 free Geometry: Circles questions in JEE Advanced Mathematics. Open each question or exercise to read its content, reference answers and available explanation.

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Geometry: Circles questions and exercises with answers

Showing questions 1–2 of 2.

Events $A$ and $B$ are mutually exclusive with $P(A) = 0.4$ and $P(B) = 0.3$. Find $P(A \cup B)$.Read the answer and worked solution →Events $A$ and $B$ are mutually exclusive with $P(A) = 0.4$ and $P(B) = 0.3$. Find $P(A \cup B)$.Read the answer and worked solution →

Worked practice questions

Events $A$ and $B$ are mutually exclusive with $P(A) = 0.4$ and $P(B) = 0.3$. Find $P(A \cup B)$.

  1. $0.12$
  2. $0.70$
  3. $1.00$
  4. $0.58$
Show answer and solution

Correct answer: B. $0.70$

Mutually exclusive events cannot occur together, so $P(A \cap B) = 0$. The general addition rule simplifies: $P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.4 + 0.3 - 0 = 0.7$. Option A ($0.12$) is the product $P(A) \cdot P(B)$, which is the joint probability for *independent* events, not mutually exclusive ones. **The distinction matters:** mutually exclusive means $P(A \cap B) = 0$ independent means $P(A \cap B) = P(A) \cdot P(B)$. These two assumptions cannot both hold unless one event has zero probability.

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Events $A$ and $B$ are mutually exclusive with $P(A) = 0.4$ and $P(B) = 0.3$. Find $P(A \cup B)$.

  1. $0.12$
  2. $0.70$
  3. $1.00$
  4. $0.58$
Show answer and solution

Correct answer: B. $0.70$

Mutually exclusive events cannot occur together, so $P(A \cap B) = 0$. The general addition rule simplifies: $P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.4 + 0.3 - 0 = 0.7$. Option A ($0.12$) is the product $P(A) \cdot P(B)$, which is the joint probability for *independent* events, not mutually exclusive ones. **The distinction matters:** mutually exclusive means $P(A \cap B) = 0$ independent means $P(A \cap B) = P(A) \cdot P(B)$. These two assumptions cannot both hold unless one event has zero probability.

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