Question

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

Answer 1

Hint: Here, we have to find whether the given two events are equally likely or not equally likely, independent or not independent and mutually exclusive by using the rule of complementary events. The set of outcomes from an experiment is known as an Event.
Formula used: We will use the following formulas:
1.Rule of Complementary events : \[P\left( A \right) + P\left( {\overline A } \right) = 1\], where \[A\] is an event and \[\overline A \] is a complementary event.
2.\[P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)\] where \[A\] and \[B\] are two events.
3.Independent event: \[P\left( {A \cap B} \right) = P\left( A \right) \cdot P\left( B \right)\]
4.Equally likely event: \[P\left( A \right) = P\left( B \right)\]
5.Mutually Exclusive event: \[P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right)\] and \[P\left( {A \cap B} \right) = 0\]
Complete step-by-step answer:
 Let \[A\] and \[B\] be two events.
Now, by using the rule of complementary events for the event \[A\] , we get
\[P\left( A \right) + P\left( {\overline A } \right) = 1\]
Substituting the value of \[P\left( A \right)\], we get
\[ \Rightarrow P\left( A \right) + \dfrac{1}{4} = 1\]
Subtracting the like terms, we get
\[ \Rightarrow P\left( A \right) = 1 - \dfrac{1}{4}\]
Here we will be taking Least Common Multiple for the denominators 1 and 4, we get
L.C.M \[\left( {1,4} \right) = 1 \times 4 = 4\]
Now, multiplying the expression in the above equation to get the denominators equal to 4 , we get
\[ \Rightarrow P\left( A \right) = 1 \times \dfrac{4}{4} - \dfrac{1}{4} \times 1\]
\[ \Rightarrow P\left( A \right) = \dfrac{4}{4} - \dfrac{1}{4}\]
Subtracting the terms, we get
\[ \Rightarrow P\left( A \right) = \dfrac{3}{4}\] .
Now, again by using the rule of complementary event, we have
\[ \Rightarrow P\left( {\overline {A \cup B} } \right) + P\left( {A \cup B} \right) = 1\]
\[ \Rightarrow P\left( {\overline {A \cup B} } \right) = 1 - P\left( {A \cup B} \right)\]
By using the formula \[P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)\], we get
\[ \Rightarrow P\left( {\overline {A \cup B} } \right) = 1 - \left( {P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)} \right)\]
Rewriting the equation, we get
\[ \Rightarrow P\left( {\overline {A \cup B} } \right) = 1 - P\left( A \right) - P\left( B \right) + P\left( {A \cap B} \right)\]
Substituting the values of \[P\left( {\overline {A \cup B} } \right) = \dfrac{1}{6}\] , \[P\left( {A \cap B} \right) = \dfrac{1}{4}\] and \[P\left( A \right) = \dfrac{3}{4}\], we get
\[ \Rightarrow \dfrac{1}{6} = 1 - \dfrac{3}{4} - P\left( B \right) + \dfrac{1}{4}\]
Now taking LCM on the right hand side, we get
\[ \Rightarrow \dfrac{1}{6} = \dfrac{4}{4} - \dfrac{3}{4} - P\left( B \right) + \dfrac{1}{4}\]
Adding and subtracting the like terms, we get
\[ \Rightarrow \dfrac{1}{6} = \dfrac{2}{4} - P\left( B \right)\]
Rewriting the equation, we have
\[ \Rightarrow P\left( B \right) = \dfrac{2}{4} - \dfrac{1}{6}\]
Again taking LCM, we get
\[ \Rightarrow P\left( B \right) = \dfrac{2}{4} \times \dfrac{3}{3} - \dfrac{1}{6} \times \dfrac{2}{2}\]
\[ \Rightarrow P\left( B \right) = \dfrac{6}{{12}} - \dfrac{2}{{12}}\]
Subtracting the terms, we get
\[ \Rightarrow P\left( B \right) = \dfrac{4}{{12}}\]
Multiplying and dividing by 4 , we get
\[ \Rightarrow P\left( B \right) = \dfrac{1}{3}\]
Now,
\[P\left( A \right) \cdot P\left( B \right) = \dfrac{3}{4} \cdot \dfrac{1}{3} = \dfrac{1}{4}\]
As \[P\left( {A \cap B} \right) = \dfrac{1}{4}\], so \[P\left( {A \cap B} \right) = P\left( A \right) \cdot P\left( B \right)\]. Therefore, the events are independent events.
Since, \[P\left( A \right) \ne P\left( B \right)\] , so the events are not equally likely events.
Since, \[P\left( {A \cap B} \right) \ne 0\] , so the events are not mutually exclusive.
Therefore, the events \[A\] and \[B\] are Independent but not equally likely.
Hence, option D is the correct answer.

Note: To solve the question, we need to have knowledge about equally likely, independent and mutually exclusive events. Two events are said to be mutually exclusive events when both events cannot occur at the same time. When the outcomes of an experiment are equally likely to happen, they are said to be equally likely events. Independent Events are not affected by any other events. The complement of an event \[A\] is said to be the event not \[A\] .