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.)

Question

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.)

a) The first two sales on Wednesday are both for more than $ 100.
b) None of the first three sales on Wednesday is for more than $ 100

Answer 1

.7^2 = .49

(1-.7)^3 = .3^3 = .027

Answer 2

To solve these two problems, we will use the concept of conditional probability. We will calculate the probability of each event step by step.

a) The probability that the first sale is for more than $100 is given as 0.70. Let's denote this event as A1.

P(A1) = 0.70

Since the event A1 already occurred, the probability for the second sale to be for more than $100, denoted as A2, is still 0.70.

P(A2|A1) = 0.70

To find the probability of both events occurring, we multiply their individual probabilities.

P(A1 and A2) = P(A1) * P(A2|A1) = 0.70 * 0.70 = 0.49

Therefore, the probability that the first two sales on Wednesday are both for more than $100 is 0.49, or 49%.

b) The probability that the first sale is not for more than $100 is given as 1 - P(A1).

P(not A1) = 1 - P(A1) = 1 - 0.70 = 0.30

Since the events are treated as independent, the probability for the second sale also not being for more than $100, denoted as not A2, is the same as P(not A1).

P(not A2|not A1) = P(not A2) = 0.30

To find the probability of none of the first three sales being for more than $100, we multiply the individual probabilities.

P(not A1 and not A2) = P(not A1) * P(not A2|not A1) * P(not A3|not A1 and not A2)

Since each sale is independent, we multiply the probabilities for each sale:

P(not A1 and not A2) = P(not A1) * P(not A2) * P(not A3) = 0.30 * 0.30 * 0.30 = 0.027

Therefore, the probability that none of the first three sales on Wednesday is for more than $100 is 0.027, or 2.7%.