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

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:

& \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{aligned}$$ So, $$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{aligned} & \dfrac{1}{4}=1-P(A) \\\ & \Rightarrow P(A)=1-\dfrac{1}{4} \\\ & \Rightarrow P(A)=\dfrac{4-1}{4}=\dfrac{3}{4} \\\ \end{aligned}$$ Hence, $$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{aligned} & P(B)=\dfrac{20+6-18}{24} \\\ & \Rightarrow P(B)=\dfrac{8}{24} \\\ & \Rightarrow P(B)=\dfrac{1}{3} \\\ \end{aligned}$$ Consider $$\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.