Assume that in family ,each child is equally likely to be a boy or a girl . A family with three children is chosen at random . The probalility that the eldest child is girl given that the family has atleast one girl is

Options

A. `1/2`
B. `1/3`
C. `2/3`
D. `4/7`

Answer

To solve the problem, we need to find the probability that the eldest child is a girl given that the family has at least one girl. We will denote the events as follows:

- Let \( E \) be the event that the eldest child is a girl.
- Let \( A \) be the event that there is at least one girl in the family.

We want to find \( P(E | A) \), which is the conditional probability of \( E \) given \( A \). We can use the formula for conditional probability:

\[ P(E | A) = \frac{P(E \cap A)}{P(A)} \]

### Step 1: Calculate \( P(A) \)

To find \( P(A) \), we first need to calculate the probability of having at least one girl in a family of three children. The total number of possible combinations of children (boys and girls) is \( 2^3 = 8 \). The combinations are:

1. BBB
2. BBG
3. BGB
4. BGG
5. GBB
6. GBG
7. GGB
8. GGG

Out of these, the only combination that does not have at least one girl is "BBB". Therefore, the probability of having at least one girl is:

\[ P(A) = 1 - P(\text{no girls}) = 1 - P(BBB) = 1 - \frac{1}{8} = \frac{7}{8} \]

### Step 2: Calculate \( P(E \cap A) \)

Next, we need to find \( P(E \cap A) \), which is the probability that the eldest child is a girl and there is at least one girl in the family. If the eldest child is a girl, the possible combinations of the three children are:

1. GBB
2. GBG
3. GGB
4. GGG

All of these combinations have at least one girl (the eldest child). The probability of each combination occurring is \( \frac{1}{8} \), so:

\[ P(E \cap A) = P(GBB) + P(GBG) + P(GGB) + P(GGG) = \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} = \frac{4}{8} = \frac{1}{2} \]

### Step 3: Calculate \( P(E | A) \)

Now we can substitute \( P(E \cap A) \) and \( P(A) \) into the conditional probability formula:

\[ P(E | A) = \frac{P(E \cap A)}{P(A)} = \frac{\frac{1}{2}}{\frac{7}{8}} = \frac{1}{2} \times \frac{8}{7} = \frac{4}{7} \]

Thus, the probability that the eldest child is a girl given that the family has at least one girl is:

\[ \boxed{\frac{4}{7}} \]