Therese Felser manages a grocery warehouse which encourages volume shopping on the part of its customers. Therese has discovered that, on any given weekday 70 percent of the customer sales amount to more than $ 100. That is any given sale on such a day has a probability of 0.70 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.) The first two sales on Wednesday are both for more than $ 100. None of the first three sales on Wednesday is for more than $ 100.

To calculate the probabilities, we need to use conditional probability and the concept of independence.

Let's break down the problem into two parts:

Part 1: Finding the probability of the first two sales being more than $100
The probability that the first sale on Wednesday is more than $100 is given as 0.70.
Assuming independence, the probability that the second sale on Wednesday is more than $100 is also 0.70.
To find the probability of both events happening, we multiply these two probabilities together:
P(First two sales > $100) = P(First sale > $100) * P(Second sale > $100)
= 0.70 * 0.70
= 0.49 or 49%

Part 2: Finding the probability of the first three sales being less than or equal to $100
The probability that any given sale on Wednesday is less than or equal to $100 is the complement of the probability of it being more than $100. So, for any given sale, the probability is 1 - 0.70 = 0.30.
Again, assuming independence, the probability that all three sales are less than or equal to $100 is:
P(First three sales ≤ $100) = P(First sale ≤ $100) * P(Second sale ≤ $100) * P(Third sale ≤ $100)
= 0.30 * 0.30 * 0.30
= 0.027 or 2.7%

So, the probability that the first two sales are more than $100 and the first three sales are less than or equal to $100 is 0.49% (or 0.0049).