Question

Let A and B be two events such that \[P\left( \overline{A\cup B} \right)=\dfrac{1}{6}\text{ and P}\left( A\cap B \right)=\dfrac{1}{4}\text{ and P}\left( \overline{A} \right)=\dfrac{1}{4}\] where $\overline{A}$ stands for the complement of the event A. Then, the events A and B are
A. Mutually exclusive and independent
B. Equally likely but not independent
C. Independent but not equally likely
D. Independent and equally likely

Answer 1

Hint: We have a formula relation \[P\left( A\cup B \right)\text{ to P(A) and P(B)}\] it is given as \[P\left( A\cup B \right)\text{ = P(A) + P(B) - P}\left( A\cap B \right)\] First of all, we will calculate the probability of A is P(A) and probability of $A\cup B\text{ is P}\left( A\cup B \right)$ using the formula \[P\left( \overline{A} \right)=1-P\left( A \right)\] Then, we will calculate the value of P(B) using above stated formula and try to find relation between A and B.

Complete step by step answer:
To solve this question, we will first define all the terms of probability which are given as follows:
Mutually exclusive: Two events A and B are called mutually exclusive if they both simultaneously cannot occur at the same time.
Equally likely: Two events A and B are called equally likely if they have the same theoretical probability (or likelihood) of occurring. I.e. P (A) = P (B).
Independent: Two events A and B are said to be independent if
\[P\left( A\cap B \right)=P\left( A \right)\times P\left( B \right)\]
Here, we are given:
\[P\left( \overline{A\cup B} \right)=\dfrac{1}{6}\text{ and P}\left( A\cap B \right)=\dfrac{1}{4}\text{ and P}\left( \overline{A} \right)=\dfrac{1}{4}\]
Now, we have that \[P\left( \overline{A} \right)=1-P\left( A \right)\] where $\overline{A}$ is complement of event A.
\[\Rightarrow P\left( \overline{A\cup B} \right)=1-P\left( A\cup B \right)\]
Substituting value of $P\left( \overline{A\cup B} \right)$ in above, we get:
\[\begin{align}
  & \dfrac{1}{6}=1-P\left( A\cup B \right) \\
 & \Rightarrow P\left( A\cup B \right)=1-\dfrac{1}{6} \\
 & \Rightarrow P\left( A\cup B \right)=\dfrac{6-1}{5}=\dfrac{5}{6} \\
\end{align}\]
So, \[P\left( A\cup B \right)=\dfrac{5}{6}\]
Again similarly using equation (i) we have
\[P\left( \overline{A} \right)=1-P\left( A \right)\]
We are given \[P\left( \overline{A} \right)=\dfrac{1}{4}\]
Substituting this value in above, we get:
\[\begin{align}
  & \dfrac{1}{4}=1-P(A) \\
 & \Rightarrow P(A)=1-\dfrac{1}{4} \\
 & \Rightarrow P(A)=\dfrac{4-1}{4}=\dfrac{3}{4} \\
\end{align}\]
Hence, \[P(A)=\dfrac{3}{4}\]
Now, we have a formula relating $P\left( A\cup B \right)$ with P (A) and P (B) which is given as
\[P\left( A\cup B \right)\text{ = P(A) + P(B) - P}\left( A\cap B \right)\]
Substituting value of \[P\left( A\cup B \right)=\dfrac{5}{6}\text{ , P(A)=}\dfrac{3}{4}\text{ and P}\left( A\cap B \right)=\dfrac{1}{4}\] in above we get:
\[P\left( A\cup B \right)\text{ = }\dfrac{5}{6}\text{ = }\dfrac{3}{4}\text{+P(B)-}\dfrac{1}{4}\]
Taking all other terms apart from P (B) to other side, we get:
\[P(B)=\dfrac{5}{6}+\dfrac{1}{4}-\dfrac{3}{4}\]
Taking LCM of RHS and solving, we get:
\[\begin{align}
  & P(B)=\dfrac{20+6-18}{24} \\
 & \Rightarrow P(B)=\dfrac{8}{24} \\
 & \Rightarrow P(B)=\dfrac{1}{3} \\
\end{align}\]
Consider \[\text{P}\left( A\cap B \right)=\dfrac{1}{4}\text{ and P(A)=}\dfrac{3}{4}\text{ and P(B)=}\dfrac{1}{3}\]
Multiplying P (A) with P (B) we get:
\[P(A)\times P(B)=\dfrac{3}{4}\times \dfrac{1}{3}=\dfrac{1}{4}=P\left( A\cap B \right)\]
So, the value \[P\left( A\cap B \right)=P(A)\times P(B)\]
Hence, A and B are independent events.
Now, finally we will check if they are equally likely or not
\[\text{P(A)=}\dfrac{3}{4}\text{ and P(B)=}\dfrac{1}{3}\]
Clearly \[P(A)\ne P(B)\]
So, A and B are not equally likely.

So, the correct answer is “Option C”.

Note: Checking from options here, as which one is the answer is tough in this question. This is so as we are not given what exactly are events A and B. Therefore, we cannot really check if they are mutually exclusive or not. Hence, going step by step to solve would lead to better results in this question.