Probability Question 216
Question: Let A and B be two events such that $ P\overline{(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 complement of event A. Then events A and B are
[AIEEE 2005]
Options:
A) Independent but not equally likely
B) Mutually exclusive and independent
C) Equally likely and mutually exclusive
D) Equally likely but not independent
Show Answer
Answer:
Correct Answer: A
Solution:
$ P( \overline{A\cup B} )=\frac{1}{6};P( A\cap B )=\frac{1}{4} $ , $ P( {\bar{A}} )=\frac{1}{4}\Rightarrow P( A )=\frac{3}{4} $ , $ P( \overline{A\cup B} )=1-P( A\cup B )=1-P( A )-P( B )+P( A\cap B ) $
Therefore $ \frac{1}{6}=\frac{1}{4}-P( B )+\frac{1}{4} $
Therefore$ P( B )=\frac{1}{3} $ .
Since $ P( A\cap B )=P( A )P( B ) $ and $ P( A )\ne P( B ) $
A and B are independent but not equally likely.
BETA
