In a family with 7 children what is the probability of having 3 boys and then 4 girls, in that order? (Exclude multiple births and assume all outcomes are equally likely).
  

In a family with 7 children, what is the probability of having 3 boys and 4 girls, in any order? (Exclude multiple births and assume all outcomes are equally likely).
  

### 1. Probability of having **3 boys and then 4 girls**, in that specific order:

Each child has a probability of 1/2 (50%) of being either a boy or a girl, and since the events are independent, the probability of having a sequence of specific genders can be calculated as the product of individual probabilities.

The probability for each child:
- Boy: \( \frac{1}{2} \)
- Girl: \( \frac{1}{2} \)

So, for the sequence of 3 boys followed by 4 girls, the probability is:

\[
\text{P(3 boys, then 4 girls)} = \left( \frac{1}{2} \right)^3 \times \left( \frac{1}{2} \right)^4 = \left( \frac{1}{2} \right)^7 = \frac{1}{128}
\]

Thus, the probability of having 3 boys followed by 4 girls in that exact order is \( \frac{1}{128} \).

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### 2. Probability of having **3 boys and 4 girls in any order**:

This is a binomial probability problem. We need to find the probability of getting exactly 3 boys and 4 girls, regardless of the order.

The probability of having a boy or a girl for each child is \( \frac{1}{2} \). The number of ways to arrange 3 boys and 4 girls in any order is determined by the number of combinations of choosing 3 boys from 7 children, which is represented by the binomial coefficient:

\[
\text{Number of ways to arrange 3 boys in 7 children} = \binom{7}{3} = \frac{7!}{3!(7-3)!} = 35
\]

The probability of each individual arrangement (since all are equally likely) is:

\[
\left( \frac{1}{2} \right)^7 = \frac{1}{128}
\]

Thus, the total probability of having 3 boys and 4 girls in any order is:

\[
\text{P(3 boys, 4 girls in any order)} = \binom{7}{3} \times \left( \frac{1}{2} \right)^7 = 35 \times \frac{1}{128} = \frac{35}{128}
\]

So, the probability of having 3 boys and 4 girls in any order is \( \frac{35}{128} \).

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Would you like more details or clarification on any of these steps?

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Here are 5 related questions that build on this topic:
1. What is the probability of having 4 boys and 3 girls in any order?
2. If the family had 8 children, what would be the probability of having 5 boys and 3 girls in any order?
3. How does the probability change if the likelihood of having a boy is slightly higher than a girl, say 0.52 vs. 0.48?
4. What is the probability of having all children of the same gender (all boys or all girls)?
5. How would you calculate the probability of having exactly 3 boys, but no restriction on the number of girls?

**Tip**: In binomial probability problems, the key is to determine how many ways the desired outcomes can occur, then multiply by the probability of each arrangement.

In a family with 7 children what is the probability of having 3 boys and then 4 girls, in that order? (Exclude multiple births and assume all outcomes are equally likely).

In a family with 7 children, what is the probability of having 3 boys and 4 girls, in any order? (Exclude multiple births and assume all outcomes are equally likely).

Learn how to calculate the probability of having 3 boys followed by 4 girls in a family with 7 children. This detailed explanation uses probability and combinatorics concepts, including binomial distribution and the binomial theorem. Perfect for high school students (Grades 10-12).

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