Question

Therese manages a grocery warehouse that encourages volume shopping on the part of its customers. Therese has discovered that on any given​ weekday, ​95% of the customer sales amount to more than​ $100. That​ is, any given sale on such a day has a probability of .9 of being for more than​ $100. (Actually, the conditional probabilities throughout the day would change​ slightly, depending on earlier​ sales, but this effect would be negligible for the first several sales of the​ day, so we can treat them as​ independent.) Find the probability of the following event. Exactly one of the first three sales on Wednesday is for more than​ $100

Answer 1

To find the probability of exactly one of the first three sales on a weekday being for more than $100 when the probability of such a sale is 95%, we use the binomial distribution formula. The calculation results in a probability of 0.7125%.

Step-by-step explanation:

The question asks to find the probability that exactly one of the first three sales on a given weekday is for more than $100, given that any sale on such a day has a probability of 0.95 of being for more than $100.

We can model each sale as a Bernoulli trial, where a success is a sale of more than $100. Because we want exactly one success (more than $100 sale) out of three trials (sales), we 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, and (1 - p) is the probability of failure.

In this case, k = 1 (one success), n = 3 (three trials), and p = 0.95 (probability of a sale being more than $100). The probability of failure (a sale of $100 or less) is therefore 1 - 0.95 = 0.05.

The calculation is as follows:
P(X = 1) = 3C1 * 0.95^1 * 0.05^2
= 3 * 0.95 * 0.0025
= 0.007125
The final probability is therefore 0.007125, or 0.7125%.