Let A and B be two events such that PA∪B¯=1/6, PA∩B=1/4 and PA¯=1/4, where A¯ stands for the complement of the event A. Then the events A and B are

Correct answer: (A) Independent but not equally likely.. PA∪B¯=1/6 ∴ PA∪B=5/6 PA∩B=1/4 PA¯=1/4 ∴ PA=3/4 ∴ PA∪B=PA+PB-PA∩B ∴ 5/6=3/4+PB-1/4 ∴ PB=1/3 ∴ PA·PB=1/3×3/4=1/4=PA∩B Therefore, A and B are independent and since PA≠PB they are not equally likely.

Q56 / 90 · JEE Main 2014 (06 Apr) · Mathematics

Let $A$ and $B$ be two events such that $P (\bar{A \cup B}) = \frac{1}{6}, P (A \cap B) = \frac{1}{4}$ and $P (\bar{A}) = \frac{1}{4},$ where $\bar{A}$ stands for the complement of the event $A$ . Then the events $A$ and $B$ are

Correct answer
A
Independent but not equally likely.
B
Independent and equally likely.
C
Mutually exclusive and independent.
D
Equally likely but not independent.

Solution

$P (\bar{A \cup B}) = \frac{1}{6}$

$∴ P (A \cup B) = \frac{5}{6}$

$P (A \cap B) = \frac{1}{4}$

$P (\bar{A}) = \frac{1}{4}$

$∴ P (A) = \frac{3}{4}$

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

$∴ \frac{5}{6} = \frac{3}{4} + P (B) - \frac{1}{4}$

$∴ P (B) = \frac{1}{3}$

$∴ P (A) \cdot P (B) = \frac{1}{3} \times \frac{3}{4} = \frac{1}{4} = P (A \cap B)$

Therefore, $A$ and $B$ are independent and since $P (A) \neq P (B)$ they are not equally likely.

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

Let A and B be two events such that PA∪B¯=16, PA∩B=14 and PA¯=14, where A¯ stands for the complement of the event A. Then the events A and B are

Solution

Relevant Questions

Let $A$ and $B$ be two events such that $P (\bar{A \cup B}) = \frac{1}{6}, P (A \cap B) = \frac{1}{4}$ and $P (\bar{A}) = \frac{1}{4},$ where $\bar{A}$ stands for the complement of the event $A$ . Then the events $A$ and $B$ are