Two buses A and B are scheduled to arrive at a town central bus station at noon. The probability A will be late is `1/5`. The probability that bus B will be late is `7/25`. The probability that the bus B is late given that bus A is late is `9/10`. Then the probabilities:
(a) probability that neither bus will be late on a particular date is `7/10`
(b) probability that bus A is late given that bus B is late is `9/14`

Explanation

To solve the problem step by step, we'll break it down into two parts as specified in the question.

### Given Information:
1. Probability that bus A will be late, \( P(A) = \frac{1}{5} \)
2. Probability that bus B will be late, \( P(B) = \frac{7}{25} \)
3. Probability that bus B is late given that bus A is late, \( P(B|A) = \frac{9}{10} \)

### Part (a): Probability that neither bus will be late

We need to find the probability that neither bus A nor bus B is late, denoted as \( P(A^c \cap B^c) \).

Using the formula:
\[ P(A^c \cap B^c) = 1 - P(A \cup B) \]
where \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \).

First, we need to find \( P(A \cap B) \):
Using the conditional probability:
\[ P(A \cap B) = P(B|A) \cdot P(A) = \frac{9}{10} \cdot \frac{1}{5} = \frac{9}{50} \]

Now, we can calculate \( P(A \cup B) \):
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
Substituting the values:
\[ P(A \cup B) = \frac{1}{5} + \frac{7}{25} - \frac{9}{50} \]

Converting \( \frac{1}{5} \) to a fraction with a denominator of 50:
\[ \frac{1}{5} = \frac{10}{50} \]
Converting \( \frac{7}{25} \) to a fraction with a denominator of 50:
\[ \frac{7}{25} = \frac{14}{50} \]

Now substituting:
\[ P(A \cup B) = \frac{10}{50} + \frac{14}{50} - \frac{9}{50} = \frac{15}{50} = \frac{3}{10} \]

Now substituting back to find \( P(A^c \cap B^c) \):
\[ P(A^c \cap B^c) = 1 - P(A \cup B) = 1 - \frac{3}{10} = \frac{7}{10} \]

### Part (b): Probability that bus A is late given that bus B is late

We need to find \( P(A|B) \):
Using the formula:
\[ P(A|B) = \frac{P(A \cap B)}{P(B)} \]

We already found \( P(A \cap B) = \frac{9}{50} \) and \( P(B) = \frac{7}{25} \).

Now substituting these values:
\[ P(A|B) = \frac{\frac{9}{50}}{\frac{7}{25}} = \frac{9}{50} \cdot \frac{25}{7} = \frac{9 \cdot 25}{50 \cdot 7} = \frac{225}{350} = \frac{9}{14} \]

### Final Answers:
(a) The probability that neither bus will be late is \( \frac{7}{10} \).

(b) The probability that bus A is late given that bus B is late is \( \frac{9}{14} \).