For two events A and B `P(A)=0.7, P(B)=0.5` and `P(AnnB)=0.3`. Find the values of following:
(i) `P(AuuB)` (ii) `P(A//B)`
(iii) `P(barA//barB)` (iv) `P(barB//barA)`

Explanation

To solve the problem step by step, we will use the given probabilities of events A and B, as well as their intersection.

Given:
- \( P(A) = 0.7 \)
- \( P(B) = 0.5 \)
- \( P(A \cap B) = 0.3 \)

We need to find the following probabilities:

### (i) \( P(A \cup B) \)

Using the formula for the union of two events:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]

Substituting the values:
\[ P(A \cup B) = 0.7 + 0.5 - 0.3 \]
\[ P(A \cup B) = 1.2 - 0.3 = 0.9 \]

### (ii) \( P(A | B) \)

Using the formula for conditional probability:
\[ P(A | B) = \frac{P(A \cap B)}{P(B)} \]

Substituting the values:
\[ P(A | B) = \frac{0.3}{0.5} \]
\[ P(A | B) = 0.6 \]

### (iii) \( P(\bar{A} | \bar{B}) \)

Using the formula for conditional probability:
\[ P(\bar{A} | \bar{B}) = \frac{P(\bar{A} \cap \bar{B})}{P(\bar{B})} \]

First, we need to find \( P(\bar{A} \cap \bar{B}) \):
\[ P(\bar{A} \cap \bar{B}) = 1 - P(A \cup B) \]
Substituting the value we found for \( P(A \cup B) \):
\[ P(\bar{A} \cap \bar{B}) = 1 - 0.9 = 0.1 \]

Now, we find \( P(\bar{B}) \):
\[ P(\bar{B}) = 1 - P(B) = 1 - 0.5 = 0.5 \]

Now substituting these values into the conditional probability:
\[ P(\bar{A} | \bar{B}) = \frac{0.1}{0.5} = 0.2 \]

### (iv) \( P(\bar{B} | \bar{A}) \)

Using the formula for conditional probability:
\[ P(\bar{B} | \bar{A}) = \frac{P(\bar{B} \cap \bar{A})}{P(\bar{A})} \]

First, we need to find \( P(\bar{B} \cap \bar{A}) \):
\[ P(\bar{B} \cap \bar{A}) = 1 - P(A \cup B) = 0.1 \quad \text{(as calculated previously)} \]

Now, we find \( P(\bar{A}) \):
\[ P(\bar{A}) = 1 - P(A) = 1 - 0.7 = 0.3 \]

Now substituting these values into the conditional probability:
\[ P(\bar{B} | \bar{A}) = \frac{0.1}{0.3} = \frac{1}{3} \approx 0.333 \]

### Summary of Results:
1. \( P(A \cup B) = 0.9 \)
2. \( P(A | B) = 0.6 \)
3. \( P(\bar{A} | \bar{B}) = 0.2 \)
4. \( P(\bar{B} | \bar{A}) = \frac{1}{3} \)