A certain company's IT department has found that 18% of all visitors to the company website arrive there in error. (They are actually looking for another website.) If this percentage is correct, what is the probability that, in a random sample of 6 visitors to the website, exactly 1 arrives in error?Round your answer to three decimal places.
Question
A certain company's IT department has found that 18% of all visitors to the company website arrive there in error. (They are actually looking for another website.) If this percentage is correct, what is the probability that, in a random sample of 6 visitors to the website, exactly 1 arrives in error?Round your answer to three decimal places.
Solution
This problem can be solved using the binomial probability formula, which is:
P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k))
where:
- P(X=k) is the probability of k successes in n trials
- C(n, k) is the combination of n items taken k at a time
- p is the probability of success on a single trial
- n is the number of trials
- k is the number of successes
In this case:
- n (the number of trials) is 6 (the number of visitors)
- k (the number of successes) is 1 (the number of visitors who arrive in error)
- p (the probability of success on a single trial) is 0.18 (the percentage of visitors who arrive in error)
Substituting these values into the formula gives:
P(X=1) = C(6, 1) * (0.18^1) * ((1-0.18)^(6-1))
First, calculate C(6, 1), which is the number of ways to choose 1 success from 6 trials. This is simply 6.
Next, calculate (0.18^1), which is simply 0.18.
Then, calculate ((1-0.18)^(6-1)), which is (0.82^5) = 0.37 (rounded to two decimal places).
Finally, multiply these three values together to get the probability:
P(X=1) = 6 * 0.18 * 0.37 = 0.3996
Rounding to three decimal places, the probability that exactly 1 out of 6 visitors arrives at the website in error is 0.400.
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