Consider a family with two children. Assume that each child is as likely to be a boy as it is to be a girl. Find the conditional probability that both children are boys, given that (i) the older child is a boy, (ii) at least one of the children is a boy.

Answer

To solve the problem, we will follow these steps:

### Step 1: Define the Sample Space
In a family with two children, where each child can either be a boy (B) or a girl (G), the possible combinations of children can be represented as follows:
- BB (both children are boys)
- BG (older child is a boy, younger child is a girl)
- GB (older child is a girl, younger child is a boy)
- GG (both children are girls)

Thus, the sample space \( S \) is:
\[ S = \{ BB, BG, GB, GG \} \]

### Step 2: Calculate the Conditional Probability for (i)
We need to find the conditional probability that both children are boys given that the older child is a boy.

Let:
- Event \( M \): both children are boys (BB)
- Event \( N \): older child is a boy (BB or BG)

Now, we need to find \( P(M | N) \), which is given by the formula:
\[ P(M | N) = \frac{P(M \cap N)}{P(N)} \]

#### Step 2.1: Find \( P(M \cap N) \)
The event \( M \cap N \) occurs when both children are boys and the older child is a boy. This only happens in the case of BB. Thus:
\[ P(M \cap N) = \frac{1}{4} \] (1 favorable outcome out of 4 total outcomes)

#### Step 2.2: Find \( P(N) \)
The event \( N \) occurs when the older child is a boy. This happens in the cases of BB and BG. Thus:
\[ P(N) = \frac{2}{4} = \frac{1}{2} \] (2 favorable outcomes out of 4 total outcomes)

#### Step 2.3: Calculate \( P(M | N) \)
Now substituting back into the formula:
\[ P(M | N) = \frac{P(M \cap N)}{P(N)} = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{4} \times \frac{2}{1} = \frac{1}{2} \]

### Step 3: Calculate the Conditional Probability for (ii)
Now we need to find the conditional probability that both children are boys given that at least one of the children is a boy.

Let:
- Event \( M \): both children are boys (BB)
- Event \( N \): at least one child is a boy (BB, BG, GB)

Now we need to find \( P(M | N) \) again using the same formula:
\[ P(M | N) = \frac{P(M \cap N)}{P(N)} \]

#### Step 3.1: Find \( P(M \cap N) \)
The event \( M \cap N \) occurs when both children are boys and at least one child is a boy. This occurs only in the case of BB. Thus:
\[ P(M \cap N) = \frac{1}{4} \]

#### Step 3.2: Find \( P(N) \)
The event \( N \) occurs when at least one child is a boy. This happens in the cases of BB, BG, and GB. Thus:
\[ P(N) = \frac{3}{4} \] (3 favorable outcomes out of 4 total outcomes)

#### Step 3.3: Calculate \( P(M | N) \)
Now substituting back into the formula:
\[ P(M | N) = \frac{P(M \cap N)}{P(N)} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{4} \times \frac{4}{3} = \frac{1}{3} \]

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