Question

Not all visitors to a certain company's website are customers. In fact, the website administrator estimates that about 5% 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 3 actually are looking for the website.

Answer 1

The question involves using the binomial probability formula to calculate the probability that exactly 3 out of 5 visitors to a website are looking for that website. The steps involve calculating combinations, raising probabilities to the power of the number of successes and failures, and multiplying these together.

Step-by-step explanation:

The student's question is about finding the probability that exactly 3 out of 5 randomly selected visitors to a website are actually looking for that website, given that the website administrator estimates that 95% (100% - 5%) of visitors are looking for the website. This is a binomial probability problem because we have a fixed number of independent trials (n=5), each with two possible outcomes (looking for the website or not), and a constant probability of success (p=0.95).

To calculate the probability, we'll use the binomial probability formula:

• P(X=k) = nCk × p^k × (1-p)^(n-k)

Where: P(X=k) is the probability of k successes in n trials, nCk is the number of combinations of n items taken k at a time, p is the probability of success on a single trial, and (1-p) is the probability of failure on a single trial.

Substitute the given values into the formula:

• P(X=3) = 5C3 × 0.95^3 × (1-0.95)^(5-3)

Now calculate 5C3, which is the number of ways to choose 3 successes out of 5 trials:

• 5C3 = 5! / (3! × (5-3)!)

Then the probability P(X=3) is:

• P(X=3) = 10 × 0.95^3 × 0.05^2

Perform the calculations to find the exact probability value.

Answer 2

2.14344%

Step-by-step explanation:

There are two possible outcomes here, the visitor looking for the websites (p=0.95) and the visitor looking for other websites(q=5%= 0.05). The question asking for the probability of exactly 3 visitors actually looking for the website(X=3) out of 5 visitors (n=5). There is no marking to differentiate the visitors, so the order of the visitors is not important.

The calculation will be:

P(X=x) = nCx ×(p)^x ×(q)^n-x

P(X=3) = 5C3 ×(0.95)^3 ×(0.05)^2

P(X=3) =2.14344%