Let A and B be two events such that , P(A∪B¯) = 16,P(A∩B) = 14 and , P(A¯) = 14, where A¯ stands for the complement of the event A. Then the events A and B are

Let A and B be two events such that , P \left(\right. \bar{A \cup B} \left.\right) \textrm{ }\textrm{ } = \textrm{ }\textrm{ } \frac{1}{6} , P \left(\right. A \cap B \left.\right) \textrm{ }\textrm{ } = \textrm{ }\textrm{ } \frac{1}{4} and , P \left(\right. \bar{A} \left.\right) \textrm{ }\textrm{ } = \textrm{ }\textrm{ } \frac{1}{4} , \textrm{ }\textrm{ } where \textrm{ }\textrm{ } \bar{A} \textrm{ }stands for the complement of the event A. Then the events A and B are

Engineering

Mathematics

Addition Theorem on Probability

Question

Let A and B be two events such that , $P (\bar{A \cup B}) = \frac{1}{6}, P (A \cap B) = \frac{1}{4}$ and , $P (\bar{A}) = \frac{1}{4}, where \bar{A}$ stands for the complement of the event A. Then the events A and B are

equally likely but not independent

independent but not equally likely

mutually exclusive and independent

independent and equally likely

$P (A \cup B) = \frac{5}{6}; P (A) = \frac{3}{4}$

P(A $\cup$ B) = P(A) + P(B) – P(A $\cap$ B)

$\frac{5}{6} = \frac{3}{4} + P (B) - \frac{1}{4}$

$P (B) = \frac{1}{3}$

Now, P(A $\cap$ B) = P(A).P(B)

Hence, A and B are independent events

Understanding the Problem

We are given two events A and B with the following probabilities:

$P (\bar{A \cup B}) = \frac{1}{6}$
$P (A \cap B) = \frac{1}{4}$
$P (\bar{A}) = \frac{1}{4}$

We need to determine the relationship between events A and B. Specifically, we need to check if they are independent and/or equally likely.

Step 1: Find P(A)

We know that $P (\bar{A}) = \frac{1}{4}$ . The probability of an event and its complement always sum to 1.

$P (A) + P (\bar{A}) = 1$

$P (A) = 1 - P (\bar{A}) = 1 - \frac{1}{4} = \frac{3}{4}$

Step 2: Find P(A ∪ B)

We are given $P (\bar{A \cup B}) = \frac{1}{6}$ . The complement of (A ∪ B) is the event that neither A nor B occurs.

Therefore, the probability of the union is:

$P (A \cup B) = 1 - P (\bar{A \cup B}) = 1 - \frac{1}{6} = \frac{5}{6}$

Step 3: Find P(B) using the Addition Rule

The general addition rule for two events is:

$P (A \cup B) = P (A) + P (B) - P (A \cap B)$

We know $P (A \cup B) = \frac{5}{6}$ , $P (A) = \frac{3}{4}$ , and $P (A \cap B) = \frac{1}{4}$ .

Let's plug these values into the formula to solve for P(B):

$\frac{5}{6} = \frac{3}{4} + P (B) - \frac{1}{4}$

Simplify the right side:

$\frac{3}{4} - \frac{1}{4} + P (B) = \frac{2}{4} + P (B) = \frac{1}{2} + P (B)$

Now the equation is:

$\frac{5}{6} = \frac{1}{2} + P (B)$

Subtract $\frac{1}{2}$ from both sides to isolate P(B):

$P (B) = \frac{5}{6} - \frac{1}{2}$

Convert $\frac{1}{2}$ to a fraction with denominator 6: $\frac{1}{2} = \frac{3}{6}$

$P (B) = \frac{5}{6} - \frac{3}{6} = \frac{2}{6} = \frac{1}{3}$

Step 4: Check for Independence

Two events A and B are independent if and only if:

$P (A \cap B) = P (A) \cdot P (B)$

Let's calculate the right side:

$P (A) \cdot P (B) = \frac{3}{4} \times \frac{1}{3} = \frac{3}{12} = \frac{1}{4}$

We are given that $P (A \cap B) = \frac{1}{4}$ .

Since $P (A \cap B) = P (A) \cdot P (B)$ , events A and B are independent.

Step 5: Check if they are Equally Likely

Two events are equally likely if P(A) = P(B).

We have $P (A) = \frac{3}{4}$ and $P (B) = \frac{1}{3}$ .

Since $\frac{3}{4}$

Detailed Solution

Understanding the Problem

Step 1: Find P(A)

Step 2: Find P(A ∪ B)

Step 3: Find P(B) using the Addition Rule

Step 4: Check for Independence

Step 5: Check if they are Equally Likely

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