If A and B are two independent events and `P(A)=0.2, P(B)=0.3`, then find the values of the following:
(i) `P` (A and B) (ii) `P` (A and not B)
(iii) `P` (A or B) (iv) `P` (none of A and B)

Explanation

To solve the problem step by step, we will calculate the probabilities as requested.

### Given:
- \( P(A) = 0.2 \)
- \( P(B) = 0.3 \)

### (i) Find \( P(A \text{ and } B) \) or \( P(A \cap B) \):
Since A and B are independent events, the probability of both A and B occurring is given by:
\[ P(A \cap B) = P(A) \times P(B) \]
Substituting the values:
\[ P(A \cap B) = 0.2 \times 0.3 = 0.06 \]

### (ii) Find \( P(A \text{ and not } B) \) or \( P(A \cap B') \):
To find the probability of A occurring and B not occurring, we use the independence of A and \( B' \):
\[ P(A \cap B') = P(A) \times P(B') \]
First, we need to calculate \( P(B') \):
\[ P(B') = 1 - P(B) = 1 - 0.3 = 0.7 \]
Now substituting the values:
\[ P(A \cap B') = 0.2 \times 0.7 = 0.14 \]

### (iii) Find \( P(A \text{ or } B) \) or \( P(A \cup B) \):
The probability of A or B occurring is given by:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
Substituting the values:
\[ P(A \cup B) = 0.2 + 0.3 - 0.06 = 0.5 - 0.06 = 0.44 \]

### (iv) Find \( P(\text{none of } A \text{ and } B) \) or \( P(A' \cap B') \):
The probability of neither A nor B occurring can be calculated using the independence of \( A' \) and \( B' \):
\[ P(A' \cap B') = P(A') \times P(B') \]
We already calculated \( P(A') \):
\[ P(A') = 1 - P(A) = 1 - 0.2 = 0.8 \]
Now substituting the values:
\[ P(A' \cap B') = 0.8 \times 0.7 = 0.56 \]

### Summary of Results:
1. \( P(A \cap B) = 0.06 \)
2. \( P(A \cap B') = 0.14 \)
3. \( P(A \cup B) = 0.44 \)
4. \( P(A' \cap B') = 0.56 \)