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 check whether the given probabilities \( P(A) \) and \( P(B) \) are consistently defined, we need to verify the conditions for the intersection and union of the two events.

### Part (i)

Given:
- \( P(A) = 0.5 \)
- \( P(B) = 0.7 \)
- \( P(A \cap B) = 0.6 \)

**Step 1: Check the condition for intersection.**
We need to check if:
1. \( P(A \cap B) \leq P(A) \)
2. \( P(A \cap B) \leq P(B) \)

Calculating:
- \( P(A \cap B) = 0.6 \)
- \( P(A) = 0.5 \)
- \( P(B) = 0.7 \)

Now, check:
- \( P(A \cap B) \leq P(A) \) → \( 0.6 \leq 0.5 \) (False)
- \( P(A \cap B) \leq P(B) \) → \( 0.6 \leq 0.7 \) (True)

Since the first condition is false, we conclude that the probabilities are not consistently defined.

### Part (ii)

Given:
- \( P(A) = 0.5 \)
- \( P(B) = 0.4 \)
- \( P(A \cup B) = 0.8 \)

**Step 1: Check the condition for union.**
We need to check if:
1. \( P(A \cup B) \leq P(A) + P(B) \)

Calculating:
- \( P(A) + P(B) = 0.5 + 0.4 = 0.9 \)

Now, check:
- \( P(A \cup B) \leq P(A) + P(B) \) → \( 0.8 \leq 0.9 \) (True)

**Step 2: Find \( P(A \cap B) \) using the formula for union.**
Using the formula:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
Substituting the known values:
\[ 0.8 = 0.5 + 0.4 - P(A \cap B) \]
This simplifies to:
\[ 0.8 = 0.9 - P(A \cap B) \]
Rearranging gives:
\[ P(A \cap B) = 0.9 - 0.8 = 0.1 \]

**Step 3: Check the condition for intersection.**
We need to check if:
1. \( P(A \cap B) \leq P(A) \)
2. \( P(A \cap B) \leq P(B) \)

Calculating:
- \( P(A \cap B) = 0.1 \)
- \( P(A) = 0.5 \)
- \( P(B) = 0.4 \)

Now, check:
- \( P(A \cap B) \leq P(A) \) → \( 0.1 \leq 0.5 \) (True)
- \( P(A \cap B) \leq P(B) \) → \( 0.1 \leq 0.4 \) (True)

Since both conditions are satisfied, we conclude that the probabilities are consistently defined.

### Summary of Results:
1. For (i): Not consistently defined.
2. For (ii): Consistently defined.