Check whether the following probabilities `P(A)` and `P(B)` are consistently defined
(i) `P(A)=0.5,P(B)=0.7, P(AnnB)=0.6`
(ii) `P(A)=0.5, P(B)=0.4, P(AuuB)=0.8`

Explanation

To determine whether the probabilities \( P(A) \) and \( P(B) \) are consistently defined, we need to check the conditions for consistency for both parts of the question.

### Part (i): \( P(A) = 0.5, P(B) = 0.7, P(A \cap B) = 0.6 \)

1. **Check the first condition**:
\[ P(A \cap B) \leq P(A) \]
Here, \( P(A \cap B) = 0.6 \) and \( P(A) = 0.5 \).
\[ 0.6 \leq 0.5 \quad \text{(This is false)} \]

2. **Check the second condition**:
\[ P(A \cap B) \leq P(B) \]
Here, \( P(B) = 0.7 \).
\[ 0.6 \leq 0.7 \quad \text{(This is true)} \]

Since the first condition is not satisfied, we conclude that the probabilities \( P(A) \) and \( P(B) \) are **not consistently defined**.

### Part (ii): \( P(A) = 0.5, P(B) = 0.4, P(A \cup B) = 0.8 \)

1. **Use the formula for union**:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
Plugging in the values:
\[ 0.8 = 0.5 + 0.4 - P(A \cap B) \]
Simplifying:
\[ 0.8 = 0.9 - P(A \cap B) \]
Rearranging gives:
\[ P(A \cap B) = 0.9 - 0.8 = 0.1 \]

2. **Check the first condition**:
\[ P(A \cap B) \leq P(A) \]
Here, \( P(A \cap B) = 0.1 \) and \( P(A) = 0.5 \).
\[ 0.1 \leq 0.5 \quad \text{(This is true)} \]

3. **Check the second condition**:
\[ P(A \cap B) \leq P(B) \]
Here, \( P(B) = 0.4 \).
\[ 0.1 \leq 0.4 \quad \text{(This is true)} \]

Since both conditions are satisfied, we conclude that the probabilities \( P(A) \) and \( P(B) \) are **consistently defined**.

### Summary of Results:
- Part (i): Not consistent
- Part (ii): Consistent