Question

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

Answer 1

Hint: Here, using formula find P(A) and P(B) and check whether they are equal or not, if P(A) = P(B), then events A and B are equally likely, and if they are not equal then they are not equally likely. To check whether A and B are independent of not , find multiply P(A) and P(B) and compare result with P(A∩B), if both are equal they A and B are independent else not independent.

Complete step-by-step answer:
\[P\left( {\bar A} \right) = \dfrac{1}{4}\] ⇒ \[P\left( A \right) = 1 - \dfrac{1}{4} = \dfrac{3}{4}\]
\[P\overline {(A \cup B)} = \dfrac{1}{6}\] ⇒ \[P(A \cup B) = 1 - \dfrac{1}{6} = \dfrac{5}{6}\]
On simplifying
\[P\left( {A \cap B} \right) = \dfrac{1}{4}\]
Also, P(AUB) = P(A) + P(B) − P(A∩B) \[ = \dfrac{3}{4} + P\left( B \right) - \dfrac{1}{4}\]
On simplifying
\[ \Rightarrow \dfrac{5}{6} = \dfrac{1}{2} + P\left( B \right)\]
\[ \Rightarrow P\left( B \right) = \dfrac{2}{6} = \dfrac{1}{3}\]
We have,
P(A) × P(B) = \[\dfrac{1}{3} \times \dfrac{3}{4} = \dfrac{1}{4} = P\left( {A \cap B} \right)\]
Since P(A) × P(B) = P(A∩B)
Therefore, A and B are independent but not equally likely.

So, the correct answer is “Option A”.

Note: In these types of questions, always use formulas to find the probability of each event separately.
Important point:
Let set A be an event in a sample space S, then complement of A is the set of all sample points, which are not in A and it is denoted by A′ or A .
P(AUB) means probability of occurrence of either A or B.
P(A∩B) means probability of occurrence of either A and B i.e., occurrence of both A and B.
Independent Events: Two events E and F are said to be independent if probability of occurrence of one of the events is not affected by that of the other. For any two independent events, we have
P(E ∩ F) = P(E) . P(F).
Dependent Event: Two events E and F are said to be dependent if probability of occurrence of one of the events gets affected by that of the other. For any two independent events, we have
P(E ∩ F) ≠ P(E) . P(F).