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The problem provided involves calculating the probability that a city will have a satisfactory water supply, given the events related to two reservoirs (a and b) exceeding their normal capacities. 

Let’s break down the provided information and proceed with solving it:

### Given Information:
1. **\( A \)**: Event that the water in reservoir \( a \) exceeds its normal capacity.
2. **\( B \)**: Event that the water in reservoir \( b \) exceeds its normal capacity.

Probabilities provided:
- \( P(B) = 0.8 \)
- \( P(AB) = 0.6 \) (i.e., the probability that both \( a \) and \( b \) exceed their capacities)
- \( P(A \mid \bar{B}) = 0.7 \) (i.e., the conditional probability that \( a \) exceeds its capacity given that \( b \) does not exceed its capacity)

Additional probabilities for satisfactory water supply conditions:
- The probability that the city will have satisfactory supply if **only one** reservoir exceeds capacity: \( 0.7 \)
- The probability that the city will have satisfactory supply if **both reservoirs** exceed capacity: \( 0.9 \)
- The probability that the city will have satisfactory supply if **neither** reservoir exceeds capacity: \( 0.3 \)

### To Find:
The probability that the city will have satisfactory water supply.

### Solution Approach:

We will apply the Law of Total Probability by breaking down the problem into different cases where satisfactory water supply occurs:
1. **Case 1**: Both reservoirs exceed capacity.
   - The probability of this event is \( P(AB) = 0.6 \).
   - The probability of satisfactory supply given this event is 0.9.

2. **Case 2**: Only one reservoir exceeds capacity (either \( A \) or \( B \)).
   - First, for \( A \) exceeding and \( B \) not exceeding: We use the conditional probability provided \( P(A \mid \bar{B}) = 0.7 \), and calculate \( P(\bar{B}) = 1 - P(B) = 1 - 0.8 = 0.2 \). Hence, \( P(A \cap \bar{B}) = P(A \mid \bar{B}) \times P(\bar{B}) = 0.7 \times 0.2 = 0.14 \).
   - For \( B \) exceeding and \( A \) not exceeding: This is \( P(\bar{A} \cap B) = P(B) - P(AB) = 0.8 - 0.6 = 0.2 \).
   - The probability of satisfactory supply given only one exceeds is 0.7 for both cases.

3. **Case 3**: Neither reservoir exceeds capacity.
   - The probability of this event is \( P(\bar{A} \cap \bar{B}) = 1 - P(A \cup B) = 1 - [P(A) + P(B) - P(AB)] \).
   - \( P(A \cup B) = P(A) + P(B) - P(AB) = P(A) + 0.8 - 0.6 \), but \( P(A) \) can be found using the marginal distribution.
   - The probability of satisfactory supply if neither exceeds capacity is 0.3.

Let’s now compute the total probability of satisfactory water supply using this information.The probability that the city will have a satisfactory water supply is approximately **0.796**.

This differs slightly from the answer provided in the question, which is 0.764, likely due to rounding or slight differences in the assumptions or values used in the problem. However, the approach is correct and the discrepancy may stem from minor details.

Do you want further clarification or detailed steps on any part?

### Related questions:
1. How do we apply the Law of Total Probability to different cases in probability problems?
2. What is conditional probability and how does it affect the outcome in problems like this?
3. How can rounding errors affect probability calculations in complex problems?
4. Why is it important to distinguish between independent and dependent events when calculating probabilities?
5. What is the significance of using union and intersection in probability problems like this?

### Tip:
Always double-check the conditions for independence when solving probability problems to ensure the correct application of formulas.

The water supply for a city comes from two reservoirs, a and b. The probability of satisfactory water supply is calculated based on various given probabilities related to reservoirs exceeding or not exceeding their capacities.

This problem calculates the probability of a satisfactory water supply using conditional probability and the law of total probability. It involves two reservoirs, with probabilities given for exceeding normal capacity and different scenarios for water supply satisfaction. Suitable for university-level students studying probability.

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