This problem involves calculating a binomial probability.

Problem Breakdown:

• Given Information:
  • Probability that a visitor is looking for another website ($p$) = 5% = $0.05$.
  • Probability that a visitor is not looking for another website ($q = 1 - p$) = $1 - 0.05 = 0.95$.
  • Random sample size ($n$) = 5 visitors.
  • Number of successes ($k$) = 4 (we want exactly 4 visitors looking for another website).

Binomial Probability Formula:

$P(X = k) = \binom{n}{k} \cdot p^k \cdot q^{n-k}$ Where:

• $\binom{n}{k}$ = $\frac{n!}{k!(n-k)!}$ is the number of combinations.
• $p^k$ is the probability of success raised to the power $k$.
• $q^{n-k}$ is the probability of failure raised to the power $n-k$.

Step-by-Step Solution:

1. Calculate $\binom{n}{k}$: $\binom{5}{4} = \frac{5!}{4!(5-4)!} = \frac{5}{1} = 5$

2. Compute $p^k$ and $q^{n-k}$: $p^k = (0.05)^4 = 0.00000625$ $q^{n-k} = (0.95)^1 = 0.95$

3. Substitute values into the formula: $P(X = 4) = \binom{5}{4} \cdot p^4 \cdot q^{n-4}$ $P(X = 4) = 5 \cdot 0.00000625 \cdot 0.95$

4. Perform the calculations: $P(X = 4) = 5 \cdot 0.0000059375 = 0.0000296875$

5. Round to three decimal places: $P(X = 4) \approx 0.000$

Final Answer:

The probability is approximately 0.000.

Would you like a breakdown of any specific step or more examples?

Related Questions:

1. What is the probability that exactly 3 visitors are looking for other websites in the same sample?
2. What is the probability that none of the 5 visitors are looking for other websites?
3. How does increasing the sample size affect the binomial probability?
4. Can the approximation using a Poisson distribution be applied here, and if so, how?
5. How would the probability change if $p$ were 10% instead of 5%?

Tip:

When working with small probabilities and large sample sizes, be cautious about rounding too early, as small differences in intermediate steps can affect the final answer.