Probability
Independence and Conditional Probability
Grade 12
Question:
<p>Let A and B be two events such that \(P(A/B) = \frac{1}{6}\), \(P(A \cap B) = \frac{1}{4}\) and \(P(A) = \frac{1}{4}\), where \(\overline{A}\) stands for complement of event A. Then, events A and B are</p>
<p>(a) independent but not equally likely</p>
<p>(b) mutually exclusive and independent</p>
<p>(c) equally likely and mutually exclusive</p>
Step-by-Step Solution
Key Concept: Check independence using $P(A \cap B) = P(A)P(B)$ and check if events are equally likely by comparing probabilities.
To determine the relationship between events A and B, we first check for independence.
For events A and B to be independent, the condition $P(A \cap B) = P(A)P(B)$ must hold.
Given $P(A \cap B) = \frac{1}{4}$ and $P(A) = \frac{1}{4}$.
Substituting these values into the independence condition:
$$ \frac{1}{4} = \frac{1}{4} P(B) $$
Solving for $P(B)$, we find:
$$ P(B) = 1 $$
Since $P(B) = 1$ is a valid probability and satisfies the independence condition with the given values of $P(A \cap B)$ and $P(A)$, events A and B are independent.
Next, we determine if events A and B are equally likely.
Events A and B are equally likely if $P(A) = P(B)$.
We have $P(A) = \frac{1}{4}$ and we determined $P(B) = 1$.
Since $P(A) \ne P(B)$ (i.e., $\frac{1}{4} \ne 1$), events A and B are not equally likely.
Therefore, events A and B are independent but not equally likely.
Correct Answer: A