Final answer:
To find the probability that Natasha travels to Nashville given that she drives when visiting relatives, you can use a conditional probability approach, considering the probabilities of driving to each city and the overall probability of driving.
Step-by-step explanation:
The student is essentially asking to find the probability that Natasha travels to Nashville given that she drives with a certain probability when visiting relatives. We are given that the probability she drives to Nashville is (4/5), the probability she flies to St. Louis is (1/3), and the probability that she drives when visiting relatives is (13/18). This is a conditional probability problem and can be solved using Bayes' theorem or a probability tree. Let’s denote by N the event where Natasha travels to Nashville and by Drives the event where she drives. We need to find P(N | Drives), the probability that she goes to Nashville given that she drives. We can start by defining two events: D_N as the event of Natasha driving to Nashville and D_S as the event of driving to St. Louis. Since the only other option is flying, and the total probability of either driving or flying is 1, we can find the probability that Natasha drives to St. Louis as the complement of her flying there, which is 1 - (1/3) or (2/3). We know that the overall probability of driving is (13/18), which equals the sum of the probabilities of driving to each city. Therefore: P(Drives) = P(D_N) + P(D_S) = P(N) * (4/5) + P(Not N) * (2/3) = (13/18) Where P(Not N) is the probability of Natasha not going to Nashville, which is going to St. Louis in this case. We can use this equation to solve for P(N), the probability of going to Nashville, since the likelihood of driving in either case is known.