The Wilson family was one of the first to come to the U.S. They had 4 children. Assuming that the probability of a child being a girl is .5, find the probability that the Wilson family had:

at least 2 girls? 

at most 3 girls?

To solve this problem, we'll use the binomial probability formula, as the situation involves independent trials (each child) with two possible outcomes (boy or girl). The probability of a child being a girl is \( p = 0.5 \), and the number of trials is \( n = 4 \) (since the Wilson family has 4 children).

The binomial probability formula is:
\[
P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}
\]
where:
- \( n \) is the number of trials (4 children),
- \( k \) is the number of successes (girls),
- \( p = 0.5 \) is the probability of a girl.

### Part 1: Probability of at least 2 girls
This means \( P(X \geq 2) \), which is the probability of having 2, 3, or 4 girls. To calculate this, we'll first find the probabilities for \( P(X = 2) \), \( P(X = 3) \), and \( P(X = 4) \), and then sum them up.

\[
P(X = 2) = \binom{4}{2} (0.5)^2 (0.5)^{2} = \frac{4!}{2!(4-2)!} \cdot (0.5)^4 = 6 \times 0.0625 = 0.375
\]

\[
P(X = 3) = \binom{4}{3} (0.5)^3 (0.5)^1 = \frac{4!}{3!(4-3)!} \cdot (0.5)^4 = 4 \times 0.0625 = 0.25
\]

\[
P(X = 4) = \binom{4}{4} (0.5)^4 (0.5)^0 = 1 \times 0.0625 = 0.0625
\]

Now, summing these gives:
\[
P(X \geq 2) = P(X = 2) + P(X = 3) + P(X = 4) = 0.375 + 0.25 + 0.0625 = 0.6875
\]

So, the probability that the Wilson family had **at least 2 girls** is **0.6875**.

### Part 2: Probability of at most 3 girls
This means \( P(X \leq 3) \), which is the probability of having 0, 1, 2, or 3 girls. To calculate this, we'll find the probabilities for \( P(X = 0) \), \( P(X = 1) \), \( P(X = 2) \), and \( P(X = 3) \), and then sum them up.

\[
P(X = 0) = \binom{4}{0} (0.5)^0 (0.5)^4 = 1 \times 0.0625 = 0.0625
\]

\[
P(X = 1) = \binom{4}{1} (0.5)^1 (0.5)^3 = 4 \times 0.0625 = 0.25
\]

We already calculated \( P(X = 2) = 0.375 \) and \( P(X = 3) = 0.25 \) from Part 1.

Now, summing these gives:
\[
P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0625 + 0.25 + 0.375 + 0.25 = 0.9375
\]

So, the probability that the Wilson family had **at most 3 girls** is **0.9375**.

Would you like more details on this method or any related concept?

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Here are 5 related questions for further exploration:
1. What is the probability that the Wilson family had exactly 2 girls?
2. How does the binomial distribution change if the probability of having a girl is 0.4 instead of 0.5?
3. What is the expected number of girls in the Wilson family?
4. How would you calculate the variance and standard deviation of the number of girls?
5. How would the probability change if the Wilson family had 5 children instead of 4?

**Tip**: The binomial distribution is useful when dealing with independent trials with two outcomes—success (girl) and failure (boy)—making it ideal for problems like this.

The Wilson family was one of the first to come to the U.S. They had 4 children. Assuming that the probability of a child being a girl is 0.5, find the probability that the Wilson family had: at least 2 girls? at most 3 girls?

Explore how to calculate the probability that a family with 4 children has at least 2 girls and at most 3 girls using binomial probability. Learn key probability concepts such as the binomial distribution and the combinations formula. This solution is ideal for high school students in Grades 9-12.

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