Events $A$ and $B$ are such that $P(A)=\frac{1}{2}$,$P(B)=\frac{7}{12}$ and $P(\text{not } A \text{ or not } B)=\frac{1}{4}$. State whether $A$ and $B$ are independent?
(B) It is given that $P(A)=\frac{1}{2}$,$P(B)=\frac{7}{12}$ and $P(A' \cup B')=\frac{1}{4}$.
Using De Morgan's Law,$A' \cup B' = (A \cap B)'$.
So,$P((A \cap B)') = \frac{1}{4}$.
Since $P(E') = 1 - P(E)$,we have $1 - P(A \cap B) = \frac{1}{4}$.
Therefore,$P(A \cap B) = 1 - \frac{1}{4} = \frac{3}{4}$.
Now,calculate $P(A) \times P(B) = \frac{1}{2} \times \frac{7}{12} = \frac{7}{24}$.
Since $P(A \cap B) = \frac{3}{4}$ and $P(A) \times P(B) = \frac{7}{24}$,we observe that $P(A \cap B) \neq P(A) \times P(B)$.
Thus,events $A$ and $B$ are not independent.
Using De Morgan's Law,$A' \cup B' = (A \cap B)'$.
So,$P((A \cap B)') = \frac{1}{4}$.
Since $P(E') = 1 - P(E)$,we have $1 - P(A \cap B) = \frac{1}{4}$.
Therefore,$P(A \cap B) = 1 - \frac{1}{4} = \frac{3}{4}$.
Now,calculate $P(A) \times P(B) = \frac{1}{2} \times \frac{7}{12} = \frac{7}{24}$.
Since $P(A \cap B) = \frac{3}{4}$ and $P(A) \times P(B) = \frac{7}{24}$,we observe that $P(A \cap B) \neq P(A) \times P(B)$.
Thus,events $A$ and $B$ are not independent.
Explore More
Similar Questions
In a certain town,$40\%$ of the people have brown hair,$25\%$ have brown eyes,and $15\%$ have both brown hair and brown eyes. If a person selected at random from the town has brown hair,what is the probability that they also have brown eyes?
Easy
View SolutionLet $A, B, C$ be pairwise independent events with $P(C) > 0$ and $P(A \cap B \cap C) = 0$. Then $P(A' \cap B' | C) = $
DifficultAIEEE 2011
View SolutionIf $A$ and $B$ are two independent events such that $P(A^{\prime}) = 0.75$,$P(A \cup B) = 0.65$ and $P(B) = p$,then the value of $p$ is
Bag $P$ contains $3$ white, $2$ red, $5$ blue balls and bag $Q$ contains $2$ white, $3$ red, $5$ blue balls. $A$ ball is chosen at random from $P$ and is placed in $Q$. If a ball is chosen from bag $Q$ at random, then the probability that it is a red ball is
If two events $A$ and $B$ are such that $P(A^c) = 0.3$,$P(B) = 0.4$ and $P(A \cap B^c) = 0.5$,then $P(B | A \cup B^c)$ is equal to
MediumIIT 1994
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo