Question

If $P\left( A \right) = \dfrac{2}{3}$ , $P\left( B \right) = \dfrac{1}{2}$ and $P\left( {A \cup B} \right) = \dfrac{5}{6}$ , then events A and B are
A) Mutually exclusive
B) Independent as well as mutually exhaustive
C) Independent
D) Dependent only on A

Answer 1

Hint:For all such questions we must know the meaning and conditions of terms like mutually exclusive, independent events, etc. In this question we are given $P\left( A \right),\,P\left( B \right)$ and $P\left( {A \cup B} \right)$ . Using them we can find $P\left( {A \cap B} \right)$ from the identity: $P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)$ .
Then we check for the conditions of mutually exclusive, independent events, etc.

Complete step by step Solution:
 Independent events are those events where the occurrence of one event is not dependent on the occurrence of any other event.
Condition for independent events is: \[{\text{P(A}} \cap {\text{B) = P(A)}} \times {\text{ P(B)}}\]
Whereas two events are said to be mutually exclusive or disjoint if they both cannot occur at the same time.
Condition for mutually exclusive events is: $P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right)$
From the identity $P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)$ we get that
 $P\left( {A \cap B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cup B} \right)$
 $P\left( {A \cap B} \right) = \dfrac{2}{3} + \dfrac{1}{2} - \dfrac{5}{6}$
Solving this we get,
 $P\left( {A \cap B} \right) = \dfrac{{7 - 5}}{6}$
 $P\left( {A \cap B} \right) = \dfrac{1}{3}$
Checking for the condition of independent events i.e., \[{\text{P(A}} \cap {\text{B) = P(A)}} \times {\text{ P(B)}}\] ...(1)
[LHS = left hand side, RHS = right hand side]
LHS= \[{\text{P(A}} \cap {\text{B) }}\]
      = $\dfrac{1}{3}$ ...(2)
RHS= \[{\text{ P(A)}} \times {\text{ P(B)}}\]
       = $\dfrac{2}{3} \times \dfrac{1}{2}$
Solving further, we get
RHS = $\dfrac{1}{3}$ ...(3)
From equations (2) and (3), we can see that condition for independent events (equation (1)) is satisfied.
Hence, A and B are independent events.
Now we check for the condition of mutually exclusive events i.e.,
 $P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right)$ ...(4)
LHS= \[{\text{P(A}} \cup {\text{B) }}\]
      = $\dfrac{5}{6}$ ...(5)
RHS= \[P\left( A \right) + P\left( B \right)\]
       = $\dfrac{2}{3} + \dfrac{1}{2}$
Solving this, we get
RHS = $\dfrac{7}{6}$ ...(6)
From equations (5) and (6), we can see that condition for mutually exclusive events (equation (4)) is not satisfied.
Hence, A and B are not mutually exclusive events.

Therefore, the correct option is C.

Note:One can confuse the meanings and conditions of independent events and mutually exclusive events so this must be taken care of. In such questions, some values like $P\left( A \right),P\left( B \right)$, etc may not be given directly so the identities and concepts of probability should be known thoroughly to find those values and solve the question further.