ProbabilityHardBloom L3
Question
Events A and B belong to common sample space S and have probabilities $P(A) = P(B) = \frac{1}{3}$. It is also known that $P(\overline{A} \cap \overline{B}) = \frac{7}{18}$, then $P(A/B)$ is equal to
Options
A.$\frac{1}{18}$
B.$\frac{1}{2}$
C.$\frac{1}{3}$
D.$\frac{1}{6}$
Solution
Sol. [Key Concept: Events $A,B$ in a common sample space: use the probability bounds/axioms.]
$$P(A \cap \overline{B}) = 1 - P(A \cup B) = \frac{7}{18}$$
$$\begin{matrix} \Rightarrow & P(\text{ }A \cup \text{ }B) = \frac{11}{18} \\ \Rightarrow & P(\text{ }A) + P(\text{ }B) - P(\text{ }A \cap \text{ }B) = \frac{11}{18} \\ \Rightarrow & P(\text{ }A \cap \text{ }B) = \frac{1}{18} \\ \Rightarrow & P(\text{ }A/B) = \frac{P(\text{ }A \cap \text{ }B)}{P(\text{ }B)} = \frac{1}{6} \end{matrix}$$
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