To solve this problem, we'll use the binomial probability formula. The binomial probability formula is given by:

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

Where:

• $n$ is the number of trials (in this case, the number of visitors in the sample, which is 4).
• $k$ is the number of successes (in this case, the number of visitors actually looking for the website, which is 2).
• $p$ is the probability of success on a single trial (the probability that a visitor is actually looking for the website, which is $1 - 0.10 = 0.90$).
• $\binom{n}{k}$ is the binomial coefficient, calculated as $\frac{n!}{k!(n-k)!}$.

Step 1: Calculate the Binomial Coefficient

$\binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6$

Step 2: Calculate the Probability

$P(X = 2) = 6 \times (0.90)^2 \times (0.10)^{4-2}$

$P(X = 2) = 6 \times (0.81) \times (0.01)$

$P(X = 2) = 6 \times 0.0081 = 0.0486$

Final Answer

The probability that exactly 2 out of 4 visitors are actually looking for the website is approximately 0.049 (rounded to three decimal places).

Would you like further details or have any questions about this problem?

---

Here are some related questions you might find useful:

1. What is the probability that all four visitors are looking for the website?
2. What is the probability that none of the four visitors are looking for the website?
3. How would the probability change if the sample size were 5 instead of 4?
4. What is the expected number of visitors out of 4 who are looking for the website?
5. How does the variance of the number of visitors looking for the website change with different sample sizes?
6. What is the probability that at least one visitor is looking for the website?
7. How would the probability change if the percentage of visitors looking for other websites increased to 20%?
8. What is the cumulative probability of having 0, 1, or 2 visitors looking for the website?

Tip: Understanding the binomial distribution helps in situations where you have a fixed number of independent trials with only two possible outcomes per trial.