Answer

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Hint: For independent events, \[P(A \cap B) = P(A).P(B)\] . Hence, we find \[P(A \cap B)\] and then we find \[P(A).P(B)\] and show that they both are equal.

Complete step-by-step answer:

Independent events are events such that probability of occurrence of one of them does not affect the occurrence of the other.

Independent events A and B satisfy the relation as follows:

\[P(A \cap B) = P(A).P(B){\text{ }}..........{\text{(1)}}\]

From the figure, we can observe that the sum of probability of occurrence of the event A and probability of occurrence of event B is equal to the sum of probability of occurrence of both event A and B and probability of occurrence of event A or event B.

\[P(A) + P(B) = P(A \cap B) + P(A \cup B){\text{ }}..........{\text{(2)}}\]

The probabilities of A, B and \[A \cup B\] are given as follows:

\[P(A) = \dfrac{1}{4}\]

\[P(B) = \dfrac{1}{3}\]

\[P(A \cup B) = \dfrac{1}{2}\]

Substituting these in equation (2), we get:

\[\dfrac{1}{4} + \dfrac{1}{3} = P(A \cap B) + \dfrac{1}{2}{\text{ }}\]

Simplifying the left-hand side, we get:

\[\dfrac{{3 + 4}}{{12}} = P(A \cap B) + \dfrac{1}{2}\]

\[\dfrac{7}{{12}} = P(A \cap B) + \dfrac{1}{2}\]

Now, solving for \[P(A \cap B)\] , we get:

\[P(A \cap B) = \dfrac{7}{{12}} - \dfrac{1}{2}\]

Simplifying the right-hand side of the equation, we get:

\[P(A \cap B) = \dfrac{{7 - 6}}{{12}}\]

\[P(A \cap B) = \dfrac{1}{{12}}{\text{ }}..........{\text{(3)}}\]

Hence, we obtained the value of \[P(A \cap B)\] .

Next, we compute the value of the product of probability of A and B.

We have:

\[P(A).P(B) = \dfrac{1}{4}.\dfrac{1}{3}\]

Multiplying the right-hand side of the equation, we have:

\[P(A).P(B) = \dfrac{1}{{12}}{\text{ }}...........{\text{(4)}}\]

From, equation (3) and equation (4), we observe that both the RHS are equal, hence LHS also are equal, we have:

\[P(A).P(B) = P(A \cap B)\]

This is nothing but equation (1), satisfying the condition for independent events.

Hence, we showed that A and B are independent events.

Note: A common mistake you can make is taking \[P(A) + P(B) = 1 = P(A \cup B) + P(A \cap B)\] and proceeding to solve for \[P(A \cap B)\] , which is wrong. You can observe that \[P(A) + P(B) = \dfrac{1}{4} + \dfrac{1}{3} = \dfrac{7}{{12}} \ne 1\] . However, \[P(A) + P(B) = P(A \cup B) + P(A \cap B)\] , always holds true.

Complete step-by-step answer:

Independent events are events such that probability of occurrence of one of them does not affect the occurrence of the other.

Independent events A and B satisfy the relation as follows:

\[P(A \cap B) = P(A).P(B){\text{ }}..........{\text{(1)}}\]

From the figure, we can observe that the sum of probability of occurrence of the event A and probability of occurrence of event B is equal to the sum of probability of occurrence of both event A and B and probability of occurrence of event A or event B.

\[P(A) + P(B) = P(A \cap B) + P(A \cup B){\text{ }}..........{\text{(2)}}\]

The probabilities of A, B and \[A \cup B\] are given as follows:

\[P(A) = \dfrac{1}{4}\]

\[P(B) = \dfrac{1}{3}\]

\[P(A \cup B) = \dfrac{1}{2}\]

Substituting these in equation (2), we get:

\[\dfrac{1}{4} + \dfrac{1}{3} = P(A \cap B) + \dfrac{1}{2}{\text{ }}\]

Simplifying the left-hand side, we get:

\[\dfrac{{3 + 4}}{{12}} = P(A \cap B) + \dfrac{1}{2}\]

\[\dfrac{7}{{12}} = P(A \cap B) + \dfrac{1}{2}\]

Now, solving for \[P(A \cap B)\] , we get:

\[P(A \cap B) = \dfrac{7}{{12}} - \dfrac{1}{2}\]

Simplifying the right-hand side of the equation, we get:

\[P(A \cap B) = \dfrac{{7 - 6}}{{12}}\]

\[P(A \cap B) = \dfrac{1}{{12}}{\text{ }}..........{\text{(3)}}\]

Hence, we obtained the value of \[P(A \cap B)\] .

Next, we compute the value of the product of probability of A and B.

We have:

\[P(A).P(B) = \dfrac{1}{4}.\dfrac{1}{3}\]

Multiplying the right-hand side of the equation, we have:

\[P(A).P(B) = \dfrac{1}{{12}}{\text{ }}...........{\text{(4)}}\]

From, equation (3) and equation (4), we observe that both the RHS are equal, hence LHS also are equal, we have:

\[P(A).P(B) = P(A \cap B)\]

This is nothing but equation (1), satisfying the condition for independent events.

Hence, we showed that A and B are independent events.

Note: A common mistake you can make is taking \[P(A) + P(B) = 1 = P(A \cup B) + P(A \cap B)\] and proceeding to solve for \[P(A \cap B)\] , which is wrong. You can observe that \[P(A) + P(B) = \dfrac{1}{4} + \dfrac{1}{3} = \dfrac{7}{{12}} \ne 1\] . However, \[P(A) + P(B) = P(A \cup B) + P(A \cap B)\] , always holds true.

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