Math Problem Statement

Not all visitors to a certain company's website are customers. In fact, the website administrator estimates that about 10% of all visitors to the website are looking for other websites. Assuming that this estimate is correct, find the probability that, in a random sample of 5 visitors to the website, exactly 4 actually are looking for the website. Round your response to at least three decimal places. (If necessary, consult a list of formulas.)

Solution

This problem can be solved using the binomial probability formula. The binomial probability formula is used to find the probability of getting exactly kk successes in nn independent Bernoulli trials, each with the same probability of success pp.

The binomial probability formula is given by:

P(X=k)=(nk)pk(1p)nkP(X = k) = \binom{n}{k} p^k (1-p)^{n-k}

Where:

  • nn is the number of trials (in this case, n=5n = 5),
  • kk is the number of successes (in this case, k=4k = 4),
  • pp is the probability of success on a single trial (in this case, p=0.9p = 0.9 since 90% are looking for the website),
  • 1p1-p is the probability of failure on a single trial (in this case, 1p=0.11-p = 0.1).

First, we need to calculate (nk)\binom{n}{k}, which represents the number of combinations of nn items taken kk at a time:

(54)=5!4!(54)!=5×4×3×2×1(4×3×2×1)(1)=5\binom{5}{4} = \frac{5!}{4!(5-4)!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{(4 \times 3 \times 2 \times 1)(1)} = 5

Now, substitute the values into the binomial probability formula:

P(X=4)=5×(0.9)4×(0.1)1P(X = 4) = 5 \times (0.9)^4 \times (0.1)^1

Let's calculate this step by step:

  1. (0.9)4=0.9×0.9×0.9×0.9=0.6561(0.9)^4 = 0.9 \times 0.9 \times 0.9 \times 0.9 = 0.6561
  2. (0.1)1=0.1(0.1)^1 = 0.1
  3. Multiply them together: 0.6561×0.1=0.065610.6561 \times 0.1 = 0.06561
  4. Finally, multiply by 5: 5×0.06561=0.328055 \times 0.06561 = 0.32805

So, the probability that exactly 4 out of 5 visitors are actually looking for the website is approximately 0.328 (rounded to three decimal places).

Would you like more details or have any questions?

Here are 5 related questions to expand your understanding:

  1. What is the probability that all 5 visitors are actually looking for the website?
  2. How would the probability change if the sample size were increased to 10 visitors?
  3. What is the expected number of visitors who are actually looking for the website in a random sample of 5?
  4. How would the probability change if the percentage of visitors looking for the website increased to 95%?
  5. Can the binomial probability formula be applied if the trials are not independent?

Tip: The binomial distribution is useful when you have a fixed number of independent trials, each with the same probability of success.

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Math Problem Analysis

Mathematical Concepts

Probability
Binomial Distribution
Combinatorics

Formulas

Binomial probability formula

Theorems

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Suitable Grade Level

Advanced High School