Suppose that 45% of all babies born in a particular hospital are boys. If 7 babies born in the hospital are randomly selected, what is the probability that at most 2 of them are boys?
Carry your intermediate computations to at least four decimal places, and round your answer to two demal places. Using the binomial probability.

To solve this problem using the binomial probability formula, we need to calculate the probability that at most 2 out of 7 babies born in the hospital are boys. The binomial probability formula is:

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

where:
- \( P(X = k) \) is the probability of having exactly \( k \) successes (in this case, boys) out of \( n \) trials (total babies).
- \( \binom{n}{k} \) is the binomial coefficient, calculated as \( \frac{n!}{k!(n-k)!} \).
- \( p \) is the probability of a single success (a boy being born).
- \( n \) is the number of trials (babies selected).
- \( k \) is the number of successes (boys).

Given:
- \( p = 0.45 \) (since 45% of the babies are boys)
- \( n = 7 \) (7 babies selected)
- We want the probability that at most 2 of them are boys, so we need to find \( P(X \leq 2) \), which is the sum of the probabilities \( P(X = 0) \), \( P(X = 1) \), and \( P(X = 2) \).

Let's calculate each probability:

1. **For \( P(X = 0) \):**
\[
P(X = 0) = \binom{7}{0} (0.45)^0 (0.55)^7 = 1 \times 1 \times 0.55^7 = 0.0061
\]

2. **For \( P(X = 1) \):**
\[
P(X = 1) = \binom{7}{1} (0.45)^1 (0.55)^6 = 7 \times 0.45 \times 0.55^6 = 0.0341
\]

3. **For \( P(X = 2) \):**
\[
P(X = 2) = \binom{7}{2} (0.45)^2 (0.55)^5 = 21 \times 0.2025 \times 0.55^5 = 0.1328
\]

Now, we sum these probabilities to get \( P(X \leq 2) \):

\[
P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0061 + 0.0341 + 0.1328 = 0.1730
\]

Rounding this to two decimal places, the probability that at most 2 of the 7 babies are boys is **0.17**.

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Would you like a detailed breakdown of any part of the solution, or have any further questions?

Here are some related questions to consider:

1. What is the probability that exactly 3 of the 7 babies are boys?
2. What is the probability that all 7 babies are boys?
3. How does the probability change if the probability of a boy being born increases to 50%?
4. How does the probability change if the number of babies increases to 10?
5. What is the expected number of boys in this scenario?
6. How is the binomial distribution related to the normal distribution?
7. What is the variance of the number of boys in this scenario?
8. What would be the probability of getting at most 2 boys if only 5 babies were selected?

**Tip:** When using the binomial distribution, always make sure that the trials are independent and that the probability of success remains constant across trials.

Learn how to calculate binomial probability to find the likelihood that at most 2 out of 7 babies born in a hospital are boys. Detailed step-by-step solution with intermediate computations rounded to two decimal places. Suitable for high school students (Grades 10-12).

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