Probability Question 288
Question: If $ P(A)=\frac{1}{2},P(B)=\frac{1}{3} $ and $ P(A\cap B)=\frac{7}{12}, $ then the value of $ P({A}’\cap {B}’) $ is
Options:
A) $ \frac{7}{12} $
B) $ \frac{3}{4} $
C) $ \frac{1}{4} $
D) $ \frac{1}{6} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ P({A}’\cap {B}’)=1-P(A\cup B) $
$ =1-( \frac{1}{2}+\frac{1}{3}-\frac{7}{12} )=1-\frac{1}{4}=\frac{3}{4} $ .
BETA
