Q75 / 180 · COMEDK 2024 (Evening Shift) · Mathematics

Let $\mathrm{A}$ and ${B}$ be two events such that $P(A / B)=\dfrac{1}{2}$ and $P(B / A)=\dfrac{1}{3}$ and $P(A \cap B)=\dfrac{1}{6}$ then, which one of the following is not true?

  • A
    $P\left(A^{\prime} \cap B\right)=\dfrac{1}{6}$
  • B
    $A \text { and } B \text { are independent }$
  • Correct answer
    C
    $\mathrm{A} \text { and } \mathrm{B} \text { are not independent }$
  • D
    $P(A \cup B)=\dfrac{2}{3}$

Solution

Given $P(A|B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{1}{2}$ and $P(A \cap B) = \dfrac{1}{6}$, we have $\dfrac{1/6}{P(B)} = \dfrac{1}{2}$, which implies $P(B) = \dfrac{1}{3}$.

Given $P(B|A) = \dfrac{P(A \cap B)}{P(A)} = \dfrac{1}{3}$ and $P(A \cap B) = \dfrac{1}{6}$, we have $\dfrac{1/6}{P(A)} = \dfrac{1}{3}$, which implies $P(A) = \dfrac{1}{2}$.

Check for independence: $P(A) \times P(B) = \dfrac{1}{2} \times \dfrac{1}{3} = \dfrac{1}{6}$. Since $P(A \cap B) = \dfrac{1}{6}$, we have $P(A \cap B) = P(A)P(B)$, so $A$ and $B$ are independent events.

Evaluate $P(A' \cap B) = P(B) - P(A \cap B) = \dfrac{1}{3} - \dfrac{1}{6} = \dfrac{1}{6}$. This is true.

Evaluate $P(A \cup B) = P(A) + P(B) - P(A \cap B) = \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{6} = \dfrac{3+2-1}{6} = \dfrac{4}{6} = \dfrac{2}{3}$. This is true.

Since $A$ and $B$ are independent, the statement that $A$ and $B$ are not independent is false.

Answer: $\mathrm{A} \text { and } \mathrm{B} \text { are not independent }$

Hence, the correct answer is option: 'C'.

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