Question

Let A and B be two events such that $P\left( \overline{A\cup B} \right)=\dfrac{1}{6},P(A\cap B)=\dfrac{1}{4}$ and $P\left( \overline{A} \right)=\dfrac{1}{4}$ where $\overline{A}$ stands for the compliment 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

Hint: Here, first we have to find $P(A)$ using the formula $P(A)=1-P\left( {\bar{A}} \right)$ , then find $P(B)$ using the formula:$P(A\cup B)=P(A)+P(B)-P(A\cap B)$. After this, check whether $P(A).P(B)=P(A\cap B)$, to determine whether the events A and B are independent or not.

Complete step-by-step answer:
We are given that:
 $\begin{align}
  & P\left( \overline{A\cup B} \right)=\dfrac{1}{6} \\
 & P(A\cap B)=\dfrac{1}{4} \\
 & P\left( {\bar{A}} \right)=\dfrac{1}{4} \\
\end{align}$
Since, $\overline{A}$ is the complement of A, we can say that:
$\begin{align}
  & P(A)=1-P\left( {\bar{A}} \right) \\
 & P(A)=1-\dfrac{1}{4} \\
\end{align}$
 In the next step by taking the LCM we get:
$\begin{align}
  & PA)=\dfrac{4-1}{3} \\
 & P(A)=\dfrac{3}{4} \\
\end{align}$
 Similarly, we can find the value of $P(A\cup B)$. Since, $P\left( \overline{A\cup B} \right)$ is the complement of $P(A\cup B)$ we can write:
$\begin{align}
  & P(A\cup B)=1-P\left( \overline{A\cup B} \right) \\
 & P(A\cup B)=1-\dfrac{1}{6} \\
\end{align}$
Next, by taking the LCM we will get:
$\begin{align}
  & P(A\cup B)=\dfrac{6-1}{6} \\
 & P(A\cup B)=\dfrac{5}{6} \\
\end{align}$
Next, we have to find the value of $P(B)$.
Now, we know by the formula that:
$P(A\cup B)=P(A)+P(B)-P(A\cap B)$
Next, by substituting all the values of $P(A\cup B),P(A)$ and $P\left( \overline{AUB} \right)=\dfrac{1}{6},P(A\cap B)=\dfrac{1}{4}$ we obtain:
$\dfrac{5}{6}=\dfrac{3}{4}+P(B)-\dfrac{1}{4}$
Next, by taking the LCM we get:
$\begin{align}
  & \dfrac{5}{6}=P(B)+\dfrac{3-1}{4} \\
 & \dfrac{5}{6}=P(B)+\dfrac{2}{4} \\
\end{align}$
Next, by cancellation we get:
$\dfrac{5}{6}=P(B)+\dfrac{1}{2}$
Now, by taking constants to one side and $P(B)$ to the other side we will get:
$P(B)=\dfrac{5}{6}-\dfrac{1}{2}$
By taking the LCM we obtain:
$\begin{align}
  & P(B)=\dfrac{5\times 1-3\times 1}{6} \\
 & P(B)=\dfrac{5-3}{6} \\
 & P(B)=\dfrac{2}{6} \\
\end{align}$
Now, by cancellation we get:
$P(B)=\dfrac{1}{3}$
 Now, let us find the value of $P(A).P(B)$.
$P(A).P(B)=\dfrac{3}{4}\times \dfrac{1}{3}$
By cancellation we obtain:
$P(A).P(B)=\dfrac{1}{4}$
 We also have $P(A\cap B)=\dfrac{1}{4}$.
Hence we, can write:
$P(A).P(B)=P(A\cap B)$
 Therefore, we can say that A and B are independent sets.
Since, the probabilities of $P(A)$ and $P(B)$ are not equal, we can say that they are not equally likely.
 Hence, we can say that the events A and B are independent but not equally likely.
 Therefore, the correct answer for this question is option (a).

Note: Here, if $P(A)\ne P(B)$then only the events will not be equally likely. If you are getting $P(A)=P(B)$ then the events are equally likely and it may lead to wrong answers.