Question

Let A and B be two events such that $P\left(\overline{A\cup B}\right)=\frac{1}{6},P(A\cap B)=\frac{1}{4}$ and $P\left(\overline{A}\right)=\frac{1}{4},$ where $\overline{A}$ stands for the complement of the event A. Then, the events A and B are

• A. Independent but not equally likely
• B. Independent and equally likely
• C. Mutually exclusive and independent
• D. Equally likely but not independent

Answer 1

Answer: (A)

Independent but not equally likely

$P\left(\overline{A\cup B}\right)=\left(\frac{1}{6}\right)1-P(A\cup B)=\left(\frac{1}{6}\right)\therefore P(A\cup B)=1-\left(\frac{1}{6}\right)=\left(\frac{5}{6}\right)thenP(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)\left(\frac{5}{6}\right)=(1-(\frac{1}{4}\left)\right)+P\left(B\right)-\left(\frac{1}{4}\right)P\left(B\right)=\left(\frac{5}{6}\right)-1+\left(\frac{1}{2}\right)=\left(\frac{5-6+3}{6}\right)=\left(\frac{1}{3}\right)nowP\left(A\right)P\left(B\right)=(1-(\frac{1}{4}\left)\right)\times \left(\frac{1}{3}\right)=\left(\frac{3}{4}\right)\times \left(\frac{1}{3}\right)=\left(\frac{1}{4}\right)whichisequaltoP(A\cap B)soAandBareindependentbutP\left(A\right)\ne P\left(B\right)hencenotequallylikely$