If \(\rm P(A\cup B)=\dfrac{5}{6}, P(A\cap B)=\dfrac{1}{3}\:and\:P(\bar A)=\dfrac{1}{2}\) , then which of the following is/are correct? 1. A and B are independent events. 2. A and B are mutually exclusive events. Select the correct answer using the code given below.
1 only
The problem provides us with several probabilities related to two events, A and B, and asks us to determine if they are independent or mutually exclusive.
We are given:
We need to evaluate two statements:
First, let's find the probability of event A, \( \text{P(A)} \), using the probability of its complement:
\( \text{P(A)} = 1 - \text{P(\bar A)} \)
\( \text{P(A)} = 1 - \dfrac{1}{2} = \dfrac{1}{2} \)
Next, we can find the probability of event B, \( \text{P(B)} \), using the formula for the probability of the union of two events:
\( \text{P(A \cup B)} = \text{P(A)} + \text{P(B)} - \text{P(A \cap B)} \)
Substitute the given values into the formula:
\( \dfrac{5}{6} = \dfrac{1}{2} + \text{P(B)} - \dfrac{1}{3} \)
Now, let's solve for \( \text{P(B)} \):
\( \text{P(B)} = \dfrac{5}{6} - \dfrac{1}{2} + \dfrac{1}{3} \)
To add and subtract these fractions, we find a common denominator, which is 6:
\( \text{P(B)} = \dfrac{5}{6} - \dfrac{1 \times 3}{2 \times 3} + \dfrac{1 \times 2}{3 \times 2} \)
\( \text{P(B)} = \dfrac{5}{6} - \dfrac{3}{6} + \dfrac{2}{6} \)
\( \text{P(B)} = \dfrac{5 - 3 + 2}{6} \)
\( \text{P(B)} = \dfrac{4}{6} = \dfrac{2}{3} \)
So, we have \( \text{P(A)} = \dfrac{1}{2} \) and \( \text{P(B)} = \dfrac{2}{3} \).
Two events A and B are mutually exclusive if they cannot occur at the same time. This means their intersection is an empty set, and the probability of their intersection is 0.
Condition for mutually exclusive events: \( \text{P(A \cap B)} = 0 \)
From the problem statement, we are given \( \text{P(A \cap B)} = \dfrac{1}{3} \).
Since \( \dfrac{1}{3} \neq 0 \), events A and B are not mutually exclusive.
Therefore, statement 2 is incorrect.
Two events A and B are independent if the occurrence of one does not affect the probability of the other. Mathematically, this condition is expressed as:
Condition for independent events: \( \text{P(A \cap B)} = \text{P(A)} \times \text{P(B)} \)
We have calculated \( \text{P(A)} = \dfrac{1}{2} \) and \( \text{P(B)} = \dfrac{2}{3} \). We are given \( \text{P(A \cap B)} = \dfrac{1}{3} \).
Let's calculate the product \( \text{P(A)} \times \text{P(B)} \):
\( \text{P(A)} \times \text{P(B)} = \dfrac{1}{2} \times \dfrac{2}{3} = \dfrac{1 \times 2}{2 \times 3} = \dfrac{2}{6} = \dfrac{1}{3} \)
Now, compare this product with \( \text{P(A \cap B)} \):
\( \text{P(A \cap B)} = \dfrac{1}{3} \)
\( \text{P(A)} \times \text{P(B)} = \dfrac{1}{3} \)
Since \( \text{P(A \cap B)} = \text{P(A)} \times \text{P(B)} \), the condition for independent events is satisfied.
Therefore, statement 1 is correct.
Based on our analysis:
Thus, only statement 1 is correct.
| Statement | Condition | Check | Result |
|---|---|---|---|
| A and B are independent | \( \text{P(A \cap B)} = \text{P(A)} \times \text{P(B)} \) | \( \dfrac{1}{3} = \dfrac{1}{2} \times \dfrac{2}{3} = \dfrac{1}{3} \) | True |
| A and B are mutually exclusive | \( \text{P(A \cap B)} = 0 \) | \( \dfrac{1}{3} = 0 \) | False |
| Concept | Notation | Explanation | Formula/Condition |
|---|---|---|---|
| Probability of Event A | \( \text{P(A)} \) | Likelihood of event A occurring. | \( 0 \le \text{P(A)} \le 1 \) |
| Complement of A | \( \text{\bar A} \) | The event that A does not occur. | \( \text{P(\bar A)} = 1 - \text{P(A)} \) |
| Union of A and B | \( \text{A \cup B} \) | The event that A or B or both occur. | \( \text{P(A \cup B)} = \text{P(A)} + \text{P(B)} - \text{P(A \cap B)} \) |
| Intersection of A and B | \( \text{A \cap B} \) | The event that both A and B occur. | \( \text{P(A \cap B)} \) |
| Mutually Exclusive Events | A and B | Events that cannot occur at the same time. | \( \text{P(A \cap B)} = 0 \) |
| Independent Events | A and B | Events where the occurrence of one does not affect the other. | \( \text{P(A \cap B)} = \text{P(A)} \times \text{P(B)} \) |
Understanding the relationship between different types of events is fundamental in probability. While mutually exclusive events cannot happen together, independent events have no influence on each other's probabilities. It's important not to confuse these two concepts; in fact, if two events A and B have non-zero probabilities (\( \text{P(A)} > 0 \) and \( \text{P(B)} > 0 \)), they cannot be both mutually exclusive and independent. If they are mutually exclusive, \( \text{P(A \cap B)} = 0 \). If they are independent, \( \text{P(A \cap B)} = \text{P(A)} \times \text{P(B)} \). If both probabilities are non-zero, \( \text{P(A)} \times \text{P(B)} > 0 \), which contradicts \( \text{P(A \cap B)} = 0 \).
This problem demonstrates how to use given probability values to deduce the nature of the relationship between events A and B.
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