Question

A camping store's customer service department receives a variety of emails every day. The head of the department explains that 16% of daily emails are from customers who have lost parts to their tents and are requesting replacements. Assuming this is true, if the customer service department randomly selects 3 emails to read before lunch today, what is the probability that exactly 1 of the emails is from a customer requesting a tent part replacement? Write your answer as a decimal rounded to the nearest thousandth.

Answer 1

Explanation:

To find the probability that exactly 1 of the 3 randomly selected emails is from a customer requesting a tent part replacement, we can use the binomial probability formula. In this case, we have a binomial distribution with the following parameters:

n (the number of trials) = 3 (since they are selecting 3 emails).

p (the probability of success on each trial) = 16% or 0.16 (since 16% of emails are from customers requesting tent part replacements).

x (the number of successes we're interested in) = 1 (exactly 1 email from a customer requesting a tent part replacement).

The binomial probability formula is:

P(X = x) = (n choose x) * p^x * (1 - p)^(n - x)

Where "n choose x" is the binomial coefficient, which can be calculated as:

(n choose x) = n! / (x!(n - x)!)

Let's calculate the probability:

(n choose x) = 3! / (1!(3 - 1)!) = 3! / (1! * 2!) = 3

Now, plug the values into the binomial probability formula:

P(X = 1) = 3 * (0.16)^1 * (1 - 0.16)^(3 - 1)

P(X = 1) = 3 * 0.16 * 0.84^2

P(X = 1) = 0.16128

So, the probability that exactly 1 of the emails is from a customer requesting a tent part replacement is approximately 0.161, rounded to the nearest thousandth.