Probability Question 189
Question: For two events A and B, if $ P(A)=P( \frac{A}{B} )=\frac{1}{4} $ and $ P( \frac{B}{A} )=\frac{1}{2} $ , then
[MP PET 2003]
Options:
A) A and B are independent
B) $ P( \frac{{{A}’}}{B} )=\frac{3}{4} $
C) $ P( \frac{{{B}’}}{{{A}’}} )=\frac{1}{2} $
D) All of these
Show Answer
Answer:
Correct Answer: D
Solution:
$ P( \frac{B}{A} )=\frac{1}{2} $
$ \Rightarrow \frac{P(B\cap A)}{P(A)}=\frac{1}{2} $
$ \Rightarrow P(B\cap A)=\frac{1}{8} $
$ P( \frac{A}{B} )=\frac{1}{4} $
$ \Rightarrow \frac{P(A\cap B)}{P(B)}=\frac{1}{4} $
$ \Rightarrow P(B)=\frac{1}{2} $
$ P(A\cap B)=\frac{1}{8}=P(A).P(B) $
$ \therefore $ Events A and B are independent.
Now, $ P( \frac{{{A}’}}{B} )=\frac{P({A}’\cap B)}{P(B)}=\frac{P({A}’)P(B)}{P(B)}=\frac{3}{4} $ and $ P( \frac{B’}{A’} )=\frac{P(B’\cap A’)}{P(A’)}=\frac{P(B’)P(A’)}{P(A’)}=\frac{1}{2} $ .
BETA
