Let A and B be independent events with `P(A) = 0. 3`and `P(B) = 0. 4`Find(i) `P(AnnB)` (ii) `P(AuuB)`(iii) `P (A|B) ` (iv) `P(B|A)`

Explanation

To solve the problem step by step, we will use the properties of independent events and the relevant probability formulas.

### Given:
- \( P(A) = 0.3 \)
- \( P(B) = 0.4 \)

### (i) Find \( P(A \cap B) \) (Probability of A intersection B)

Since A and B are independent events, the probability of their intersection is given by:
\[ P(A \cap B) = P(A) \times P(B) \]
Substituting the values:
\[ P(A \cap B) = 0.3 \times 0.4 = 0.12 \]

### (ii) Find \( P(A \cup B) \) (Probability of A union B)

The probability of the union of two events is given by:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
Substituting the values we have:
\[ P(A \cup B) = 0.3 + 0.4 - 0.12 = 0.58 \]

### (iii) Find \( P(A | B) \) (Probability of A given B)

The conditional probability of A given B is calculated using the formula:
\[ P(A | B) = \frac{P(A \cap B)}{P(B)} \]
Substituting the values:
\[ P(A | B) = \frac{0.12}{0.4} = 0.3 \]

### (iv) Find \( P(B | A) \) (Probability of B given A)

Similarly, the conditional probability of B given A is given by:
\[ P(B | A) = \frac{P(A \cap B)}{P(A)} \]
Substituting the values:
\[ P(B | A) = \frac{0.12}{0.3} = 0.4 \]

### Summary of Results:
1. \( P(A \cap B) = 0.12 \)
2. \( P(A \cup B) = 0.58 \)
3. \( P(A | B) = 0.3 \)
4. \( P(B | A) = 0.4 \)