Given `P(A)=0.54, P(B)=0.69` and `P(AnnB)=0.35`
(i) `P(A' nnB')` (ii) `P(AnnB')`

Answer

To solve the problem, we need to find two probabilities: \( P(A' \cap B') \) and \( P(A \cap B') \).

### Given:
- \( P(A) = 0.54 \)
- \( P(B) = 0.69 \)
- \( P(A \cap B) = 0.35 \)

### (i) Finding \( P(A' \cap B') \)

Using De Morgan's Law, we have:
\[ P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B) \]

Now, we need to find \( P(A \cup B) \) using the formula:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]

Substituting the values:
\[ P(A \cup B) = 0.54 + 0.69 - 0.35 \]

Calculating:
\[ P(A \cup B) = 1.23 - 0.35 = 0.88 \]

Now, substituting back to find \( P(A' \cap B') \):
\[ P(A' \cap B') = 1 - P(A \cup B) = 1 - 0.88 = 0.12 \]

### (ii) Finding \( P(A \cap B') \)

To find \( P(A \cap B') \), we can use the formula:
\[ P(A \cap B') = P(A) - P(A \cap B) \]

Substituting the values:
\[ P(A \cap B') = 0.54 - 0.35 \]

Calculating:
\[ P(A \cap B') = 0.19 \]

### Final Answers:
(i) \( P(A' \cap B') = 0.12 \)
(ii) \( P(A \cap B') = 0.19 \)

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