There are two children in a family. Find the probability that both boys if it is known that:
(i) one of the children is a boy
(ii) Older child is a boy

Explanation

To solve the problem, we need to find the probability that both children in a family are boys under two different conditions. Let's break it down step by step.

### Step 1: Define the Sample Space
The possible combinations of two children can be represented as follows:
- BB (both are boys)
- BG (boy is older, girl is younger)
- GB (girl is older, boy is younger)
- GG (both are girls)

Thus, the sample space \( S \) can be represented as:
\[ S = \{ BB, BG, GB, GG \} \]

### Step 2: Condition (i) - One of the children is a boy
We need to find the probability that both children are boys given that at least one child is a boy.

**Event E**: Both children are boys (BB).
**Event F**: At least one child is a boy. The possible outcomes for this event are:
- BB
- BG
- GB

So, the event F can be represented as:
\[ F = \{ BB, BG, GB \} \]

### Step 3: Find \( P(E|F) \)
Using the formula for conditional probability:
\[ P(E|F) = \frac{P(E \cap F)}{P(F)} \]

- **Finding \( E \cap F \)**: The intersection of events E and F is just the event where both are boys, which is:
\[ E \cap F = \{ BB \} \]
Thus, \( P(E \cap F) = \frac{1}{4} \) (since there is 1 favorable outcome out of 4 total outcomes).

- **Finding \( P(F) \)**: The probability of event F (at least one boy) has 3 favorable outcomes (BB, BG, GB):
\[ P(F) = \frac{3}{4} \]

### Step 4: Calculate \( P(E|F) \)
Now substituting the values into the conditional probability formula:
\[ P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3} \]

### Step 5: Condition (ii) - Older child is a boy
Now we need to find the probability that both children are boys given that the older child is a boy.

**Event H**: The older child is a boy. The possible outcomes for this event are:
- BB (both are boys)
- BG (boy is older, girl is younger)

So, the event H can be represented as:
\[ H = \{ BB, BG \} \]

### Step 6: Find \( P(E|H) \)
Using the same formula for conditional probability:
\[ P(E|H) = \frac{P(E \cap H)}{P(H)} \]

- **Finding \( E \cap H \)**: The intersection of events E and H is:
\[ E \cap H = \{ BB \} \]
Thus, \( P(E \cap H) = \frac{1}{4} \).

- **Finding \( P(H) \)**: The probability of event H (older child is a boy) has 2 favorable outcomes (BB, BG):
\[ P(H) = \frac{2}{4} = \frac{1}{2} \]

### Step 7: Calculate \( P(E|H) \)
Now substituting the values into the conditional probability formula:
\[ P(E|H) = \frac{P(E \cap H)}{P(H)} = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{2} \]

### Final Answers
1. The probability that both children are boys given that one of the children is a boy is \( \frac{1}{3} \).
2. The probability that both children are boys given that the older child is a boy is \( \frac{1}{2} \).