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.