Suppose that 3% of the 2 million high school students who take the SAT each year receive special accommodations because of documented disabilities. Consider a random sample of 25 students who have recently taken the test. What is the probability that exactly 1 received a special accommodation?

To solve this problem, we can model it using a **binomial distribution**. The binomial distribution is appropriate because we have a fixed number of trials (students), each student either received special accommodation or did not, and the probability of receiving accommodation is constant.

The formula for the binomial probability is:

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

Where:
- \(n\) is the number of trials (students) in the sample. Here, \(n = 25\).
- \(k\) is the number of successes (students who received accommodations). We want \(k = 1\).
- \(p\) is the probability of success (receiving an accommodation). We are told that 3% of students receive accommodations, so \(p = 0.03\).
- \(1 - p\) is the probability of failure (not receiving accommodation).

Now, applying the values into the formula:

\[
P(X = 1) = \binom{25}{1} (0.03)^1 (1 - 0.03)^{25 - 1}
\]

First, we calculate each part step by step.

1. \(\binom{25}{1} = 25\)
2. \(0.03^1 = 0.03\)
3. \(1 - 0.03 = 0.97\)
4. \(0.97^{24}\)

Let me calculate this final probability.The probability that exactly 1 student out of the 25 randomly selected students received special accommodation is approximately **0.361**, or 36.1%.

Would you like further details or have any other questions?

Here are five related questions you might explore:
1. What is the probability that none of the 25 students received special accommodation?
2. What is the probability that at least 2 students received special accommodation?
3. How would the probability change if the sample size were increased to 50 students?
4. How can we calculate the expected number of students in the sample who received accommodations?
5. How would we compute the variance for the number of students receiving accommodations?

**Tip:** In binomial distributions, the expected value is calculated using \(E(X) = n \cdot p\), and variance is given by \(Var(X) = n \cdot p \cdot (1 - p)\).

Calculate the probability that exactly 1 out of 25 randomly selected students received a special accommodation for the SAT using binomial distribution. Explore step-by-step calculations with a detailed breakdown, covering essential probability concepts, binomial theorem, and related statistics.

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