In a family with 3 children, excluding multiple births, what is the probability of having 2
boys and 1 girl, in that order? Assume that a boy is as likely as a girl at each birth.
How will your answer change if any order is allowed?

### Case 1: Probability of having 2 boys and 1 girl, in that order
In this case, the specific order is required: two boys followed by one girl.

Since each child is equally likely to be a boy or a girl (with probability \( \frac{1}{2} \) for each), the probability of having two boys and one girl in that exact order (boy, boy, girl) is:

\[
P(\text{Boy, Boy, Girl}) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}
\]

So, the probability of having two boys and one girl in that specific order is \( \frac{1}{8} \).

### Case 2: Probability of having 2 boys and 1 girl, in any order
Now, if any order is allowed (not just boy, boy, girl), we need to count all the possible ways of having 2 boys and 1 girl. 

The total number of possible orders of 3 children (2 boys and 1 girl) can be found using combinations. The number of ways to arrange two boys (B) and one girl (G) is given by the number of ways to choose 2 positions out of 3 for the boys, which is:

\[
\text{Number of ways} = \binom{3}{2} = \frac{3!}{2!(3-2)!} = 3
\]

The three possible orders are:
- Boy, Boy, Girl (BBG)
- Boy, Girl, Boy (BGB)
- Girl, Boy, Boy (GBB)

Each of these has a probability of \( \frac{1}{8} \) (since each child has a \( \frac{1}{2} \) chance of being a boy or girl). Therefore, the total probability of having 2 boys and 1 girl in any order is:

\[
P(\text{2 boys, 1 girl in any order}) = 3 \times \frac{1}{8} = \frac{3}{8}
\]

### Summary:
- **In the specific order (boy, boy, girl):** Probability is \( \frac{1}{8} \).
- **In any order (2 boys, 1 girl):** Probability is \( \frac{3}{8} \).

Do you want any more details or clarifications?

### Related Questions:
1. What is the probability of having 3 boys in any order?
2. How would the probability change if the family had 4 children?
3. What is the probability of having exactly 2 girls and 1 boy in any order?
4. How do combinations and permutations differ in calculating probabilities?
5. What is the probability of having at least one girl in a family of 3 children?

### Tip:
When dealing with probabilities involving specific orders, multiplying the probabilities for each event is essential. For any order, combinations are used to count all the possibilities.

In a family with 3 children, excluding multiple births, what is the probability of having 2 boys and 1 girl, in that order? Assume that a boy is as likely as a girl at each birth. How will your answer change if any order is allowed?

Calculate the probability of having 2 boys and 1 girl in a family of 3 children, in both a specific order (boy, boy, girl) and in any order. This problem involves key probability concepts and combination formulas, making it ideal for students in Grades 6-8.

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