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A man who works in a big city owns two cars, one small and one large. Three- quarters of the time he drives the small car to work, and one-quarter of the time he takes the large car. If he takes the small car, he usually has little trouble parking and so is at work on time with probability 0.9. If he takes the large car, he is on time to work with probability 0.6. Given that he was at work on time on a particular morning, what is the probability that he drove the small car?

Short Answer

Expert verified
\( P(A|B) = \frac{27}{33} \)

Step by step solution

01

Find the probability of being on time (P(B))

We want to find P(B), which is the probability of being on time. We can use the law of total probability to find that P(B) = P(B|A) * P(A) + P(B|A') * P(A'): P(B) = 0.9 * (3/4) + 0.6 * (1/4) P(B) = 0.675 + 0.15 P(B) = 0.825
02

Calculate the probability of driving the small car and being on time (P(A and B))

Now, we need to find P(A and B), which is the probability of driving the small car and being on time. We can find this using the following formula: P(A and B) = P(B|A) * P(A) P(A and B) = 0.9 * (3/4) P(A and B) = 27/40
03

Find the probability of driving the small car given he was on time (P(A|B))

Finally, we can use the formula for conditional probability to find P(A|B): P(A|B) = P(A and B) / P(B) P(A|B) = (27/40) / 0.825 P(A|B) = 27/33 So the probability that the man drove the small car given he was on time is 27/33.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Law of Total Probability
Understanding the law of total probability is essential when analyzing situations with multiple scenarios that cover all possible outcomes. It's like looking at different paths and figuring out how likely you are to reach your destination if you were to randomly choose one of these paths.

In our example of the man with two cars, there are two 'paths' he can take each morning: driving his small car or his large car. The law of total probability tells us that if we want to calculate the probability of an event like 'being on time for work,' we must consider all the ways it could happen and add up their probabilities. Mathematically, it's expressed as: \[ P(B) = P(B|A) \times P(A) + P(B|A') \times P(A') \] where \( A \) represents taking the small car, \( A' \) the large car, and \( B \) is the event of being on time. Each term represents a different 'path' to the same outcome, being on time.
Probability of Being On Time
The assessment of someone's punctuality, in the world of probability, is a matter of determining how likely they are to be on time. This takes into account all variables that might impact their arrival.

In this scenario, the variable affecting the man's punctuality is the size of the car he chooses to drive. For example, the probability of him being on time while driving the small car is higher because it's easier for him to park. To find the overall probability of the man being on time regardless of the car choice, we used the law of total probability. By calculating \( P(B) = 0.9 \times (3/4) + 0.6 \times (1/4) \), we found that overall, there's an 82.5% chance that he will make it to work on time on any given day.
Bayes' Theorem
Bayes' theorem is a powerful tool in the field of probability that helps us update our beliefs based on new information. If we find out an event happened, Bayes' theorem helps us rewind the clock to better understand which one of our initial scenarios was most likely to bring about that result.

In our man's commute, given he arrived on time, we want to determine the probability that he drove the small car. We can do this through Bayes' theorem, which in general terms, is written as: \[ P(A|B) = \frac{P(B|A) \times P(A)}{P(B)} \] After following these steps, we find out that if the man was on time, there's a probability of \( 27/33 \) that he drove the small car. This is the revised probability and reflects how the initial belief that he drove the small car (3/4 of the time) has increased, now that we know he was indeed on time.

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Most popular questions from this chapter

Many cities regulate the number of taxi licenses, and there is a great deal of competition for both new and existing licenses. Suppose that a city has decided to sell 10 new licenses for \(\$ 25,000\) each. A lottery will be held to determine who gets the licenses, and no one may request more than 3 licenses. Twenty individuals and taxi companies have entered the lottery. Six of the 20 entries are requests for 3 licenses, nine are requests for 2 licenses, and the rest are requests for a single license. The city will select requests at random, filling as much of the request as possible. For example, if there were only one license left, any request selected would only receive this single license. a. An individual who wishes to be an independent driver has put in a request for a single license. Use simulation to approximate the probability that the request will be granted. Perform at least 20 simulated lotteries (more is better!). b. Do you think that this is a fair way of distributing licenses? Can you propose an alternative procedure for distribution?

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