Final answer:
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%.