Question

A college student frequents one of two coffee houses on campus, choosing Starbucks \(70 \%\) of the time and Peet's \(30 \%\) of the time. Regardless of where she goes, she buys a cafe mocha on \(60 \%\) of her visits. a. The next time she goes into a coffee house on campus, what is the probability that she goes to Starbucks and orders a cafe mocha? b. Are the two events in part a independent? Explain. c. If she goes into a coffee house and orders a cafe mocha, what is the probability that she is at Peet's? d. What is the probability that she goes to Starbucks or orders a cafe mocha or both?

Short answer

Answer: The probability that the student visits Starbucks or orders a cafe mocha, or both, is 0.88.

Step-by-step solution

a. Probability of visiting Starbucks and ordering a cafe mocha

To find the probability of the student visiting Starbucks and ordering a cafe mocha, we can use the conditional probability formula: P(S and M) = P(S) * P(M|S) In this case, P(S) = 0.7 and P(M|S) = 0.6. Therefore, we have: P(S and M) = 0.7 * 0.6 = 0.42

b. Are the two events independent?

To determine if the events are independent, we can check if P(S and M) = P(S) * P(M). First, we need to calculate P(M): - P(M) = P(M|S) * P(S) + P(M|P) * P(P) - P(M) = 0.6 * 0.7 + 0.6 * 0.3 - P(M) = 0.6 Now, let's check the independence condition: P(S and M) = 0.42 P(S) * P(M) = 0.7 * 0.6 = 0.42 Since P(S and M) = P(S) * P(M), the two events are independent.

c. Conditional probability of being at Peet's given a cafe mocha order

To find the probability that the student is at Peet's given that she ordered a cafe mocha, we can use the conditional probability formula: P(P|M) = P(P and M) / P(M) We already know P(M) = 0.6. We need to find P(P and M): P(P and M) = P(P) * P(M|P) P(P and M) = 0.3 * 0.6 = 0.18 Now we can calculate P(P|M): P(P|M) = P(P and M) / P(M) P(P|M) = 0.18 / 0.6 = 0.3

d. Probability of visiting Starbucks, ordering a cafe mocha, or both

To find the probability of visiting Starbucks or ordering a cafe mocha or both, we can use the formula for the probability of the union of two events: P(S or M) = P(S) + P(M) - P(S and M) We already know P(S) = 0.7, P(M) = 0.6, and P(S and M) = 0.42. Now we can calculate P(S or M): P(S or M) = 0.7 + 0.6 - 0.42 = 0.88

Key concepts

Independent Events

When we talk about independent events in probability, we're referring to scenarios where the outcome of one event does not affect the outcome of another. For instance, flipping a coin and rolling a die are independent events because the coin toss has no influence on the die roll.

To test for independence between two events, say A and B, we check whether the probability of both A and B occurring is the same as the product of their individual probabilities - that is, if P(A and B) = P(A) * P(B). If this condition is met, then A and B are independent. In our coffee house scenario, because the student chooses where to buy coffee and whether to buy a cafe mocha independently, we can say these two events are independent, which is demonstrated through the matching probabilities in step 2 of the solution.

Probability Formulas

The world of probability is rich with formulas to help us tackle a variety of questions. The fundamental one is the conditional probability formula, which tells us how likely an event is, given that another event has occurred. It's usually written as P(A|B), meaning the probability of A given B.

Another crucial one is the multiplication rule for independent events: if A and B are independent, then P(A and B) = P(A) * P(B). This is used in step 1 of our solution to find the probability of the student's coffee purchase behavior. Understanding how to apply these formulas is key to unraveling complex probability questions.

Union of Events

In probability theory, the union of events refers to occurrences where at least one of the events happens. The probability of the union of two events A and B (denoted as P(A or B)) can be calculated with the formula: P(A) + P(B) - P(A and B).

Why subtract P(A and B)? This part of the formula corrects for the fact that the intersection of A and B (when both events occur) has been counted twice. In the context of our coffee house example, we're interested in the chance that the student either goes to Starbucks, orders a cafe mocha, or does both, and this is where the formula comes into play in step 4. Understanding the union of events is essential in order to answer questions involving 'either/or' conditions, which are common in probability questions and real-world scenarios alike.