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

What is the probability that the third sale on Wednesday is for more than $ 100?

Answer: 0.30

To solve this problem, we need to find the probability that the fourth sale on Wednesday will also be for more than $100, given that the first three sales were not for more than $100.

Let's break down the information given:
- Probability of any sale on a weekday being for more than $100 is 0.70.
- 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.

Since each sale is treated as independent, the probability of a sale being for more than $100 remains the same for each sale on a given day.

To find the probability of the fourth sale on Wednesday being for more than $100, we can apply conditional probability.

Let's define the events:
A: The first three sales on Wednesday are not for more than $100.
B: The fourth sale on Wednesday is for more than $100.

We need to find P(B|A), the probability of event B occurring given that event A has occurred.

Using the definition of conditional probability:
P(B|A) = P(A and B) / P(A)

We know P(A and B) = P(B) since A and B are independent events.

P(A and B) = P(B) = 0.70 (probability of a sale being for more than $100 on any given day)

To find P(A), the probability of the first three sales on Wednesday not being for more than $100, we can use the complement rule.

P(A) = 1 - P(all three sales on Wednesday being for more than $100)

Since the first two sales are for more than $100, the probability of the third sale not being for more than $100 is:

P(third sale not being for more than $100) = 1 - 0.70 = 0.30

So,
P(A) = (0.30)^1 = 0.30

Now we can calculate P(B|A):
P(B|A) = P(A and B) / P(A)
= P(B) / P(A)
= 0.70 / 0.30
= 7/3
≈ 2.33

Therefore, the probability that the fourth sale on Wednesday will also be for more than $100, given that the first three sales were not for more than $100, is approximately 2.33/10 or 23.3%.

To analyze the probability of the first three sales on Wednesday, we need to consider two scenarios: (1) the first two sales are both for more than $100, and (2) none of the first three sales is for more than $100.

Given that any given sale on a weekday has a probability of 0.70 of being for more than $100, the probability of any given sale being for $100 or less is 1 - 0.70 = 0.30.

Scenario 1: The first two sales are both for more than $100.
Since each sale is treated as independent, the probability of the first sale being more than $100 is 0.70. The probability of the second sale also being more than $100 is also 0.70, assuming that the conditional probabilities throughout the day change negligibly.

To calculate the probability of both events happening together, we multiply the probabilities: 0.70 * 0.70 = 0.49.

Scenario 2: None of the first three sales is for more than $100.
In this case, we want the first three sales to be for $100 or less. Since the probability of any given sale being $100 or less is 0.30, we need to calculate the probability of this happening three times in a row.

To calculate the probability of this event happening three times, we multiply the probabilities: 0.30 * 0.30 * 0.30 = 0.027.

Therefore, the probability of the first two sales being both more than $100 and none of the first three sales being for more than $100 is 0.49 * 0.027 = 0.01323, or approximately 1.32%.