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This problem involves calculating probabilities using the binomial distribution, where each child has a probability of \( \frac{1}{2} \) of being a boy. The total number of trials (children) is 6, and we need to find the probabilities for two cases:

### 1) Probability of having exactly 3 boys and 3 girls:
This can be modeled using the binomial probability formula:

\[
P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}
\]

where:
- \( n = 6 \) (number of trials, i.e., 6 children),
- \( k = 3 \) (number of boys),
- \( p = \frac{1}{2} \) (probability of any child being a boy),
- \( \binom{n}{k} \) is the binomial coefficient.

First, compute \( \binom{6}{3} \):

\[
\binom{6}{3} = \frac{6!}{3!(6 - 3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20
\]

Now, the probability:

\[
P(X = 3) = 20 \times \left( \frac{1}{2} \right)^3 \times \left( \frac{1}{2} \right)^3 = 20 \times \left( \frac{1}{2} \right)^6 = 20 \times \frac{1}{64} = \frac{20}{64} = \frac{5}{16}
\]

Thus, the probability of having exactly 3 boys and 3 girls is \( \frac{5}{16} \).

### 2) Probability of having fewer boys than girls:
To have fewer boys than girls, we need to calculate the probabilities for having 0, 1, or 2 boys (since more than 3 boys would mean more boys than girls).

#### Case 1: \( X = 0 \) (0 boys)
\[
P(X = 0) = \binom{6}{0} \left( \frac{1}{2} \right)^0 \left( \frac{1}{2} \right)^6 = 1 \times \frac{1}{64} = \frac{1}{64}
\]

#### Case 2: \( X = 1 \) (1 boy)
\[
P(X = 1) = \binom{6}{1} \left( \frac{1}{2} \right)^1 \left( \frac{1}{2} \right)^5 = 6 \times \frac{1}{64} = \frac{6}{64} = \frac{3}{32}
\]

#### Case 3: \( X = 2 \) (2 boys)
\[
P(X = 2) = \binom{6}{2} \left( \frac{1}{2} \right)^2 \left( \frac{1}{2} \right)^4 = 15 \times \frac{1}{64} = \frac{15}{64}
\]

Now, add the probabilities for these cases:

\[
P(\text{fewer boys than girls}) = P(X = 0) + P(X = 1) + P(X = 2)
\]
\[
P(\text{fewer boys than girls}) = \frac{1}{64} + \frac{3}{32} + \frac{15}{64} = \frac{1}{64} + \frac{6}{64} + \frac{15}{64} = \frac{22}{64} = \frac{11}{32}
\]

Thus, the probability of having fewer boys than girls is \( \frac{11}{32} \).

---

Would you like more detailed explanations on any part of the solution or clarification? Here are some follow-up questions to deepen your understanding:

1. How is the binomial coefficient calculated?
2. What would happen if the probability of having a boy were different from \( \frac{1}{2} \)?
3. How would the results change if the family had more or fewer than 6 children?
4. How can we extend this problem to a different number of boys and girls?
5. What is the significance of symmetry in this type of binomial distribution problem?

**Tip:** Whenever you're dealing with probability in binomial experiments, look for symmetry. If \( p = 0.5 \), the number of successes and failures will often yield a balanced distribution!

A family has 6 children. Find the probability that there are 1) 3 boys and 3 girls, 2) fewer boys than girls. Assume that the probability of any particular child being a boy is 1/2.

Calculate the probability of having exactly 3 boys and 3 girls, and fewer boys than girls, in a family of 6 children. This problem uses the binomial distribution formula with a probability of 1/2 for each child being a boy. Suitable for students in Grades 10-12.

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