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Q: Statistics - Probability ( Answered ,   1 Comment )
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 Subject: Statistics - Probability Category: Science > Math Asked by: dime365-ga List Price: \$6.00 Posted: 11 Feb 2004 15:01 PST Expires: 12 Mar 2004 15:01 PST Question ID: 305906
 ```Assume that there are nine parking spaces next to one another in a parking lot. Nine cars need to be parked by an attendant. Three of the cars are expensive sports cars, three are large domestic cars and three are compacts. Assuming the attendant parks the cars at random, what is the probability that the sports cars are parked next to each other?```
 ```Good evening, First off, I assume the parking spots can be numbered 1 through 9 such that each adjacent number represents an adjacent parking spot. I'll start with the "logical" explanation, and then follow it up with the actual math. This is a probability problem that we'll solve by counting. First, we need to count the total number of ways that these 9 cars can be arranged in the 9 parking spaces. Next, we count how many of those orders have the sports cars next to each other. Since your question states that the probability of each possible order occuring is the same, we can simply divide the number of ways the sports cars can end up together by the total number of ways the cars can be arranged, and this gives the probability of the sports cars ending up together. The first part of this is easy. For parking space 1, any of the 9 cars can be parked. In the second spot, only 8 cars can (the original 9, minus the one in spot 1). In the third spot, 7 cars can be parked (the original 9, minus the car is spot 1, minus the car in spot 2), and so on. The 9 cars can be parked in 9! = 9 * 8 * 7 * . . . * 2 * 1 unique ways. If the sports cars are to be together, they must be in consecutive parking spaces. For example {1, 2, 3} or {6, 7, 8}. There are 7 such consecutive runs of 3 stalls. For each of these runs, the sports cars can be in several orders (as above, 3! different ways to order 3 cars in 3 parking stalls). Furthermore, once parking stalls are chosen for the sports cars, and the sports cars are ordered within those stalls, the other cars must be considered. For each run, and each order of sports cars, the remaining 6 cars can be placed in 6! different orders. Now we have 9! total orders that the cars can be in, and 7 * 3! * 6! of them have the three sports cars next to each other. (7 * 3! * 6!) / 9! = (3! * 7!) / 9! (3 * 7!) / 9! = 3 / (9 * 8) 3 / (9 * 8) = 1 / 24, or just over 4%. You picked a good time to ask this question - I'm taking a probability course myself, and we just covered this material. If you need it explained differently, just ask and I'll show you a different way to attack the problem. -Haversian```
 ```All the reasoning is right--just one math error. (7 * 3! * 6!) / 9! = (3! * 7!) / 9! (3 * 7!) / 9! = 3 / (9 * 8) 3 / (9 * 8) = 1 / 24, or just over 4%. In going from the first line to the second line above, a 3! got changed to a 3, so the answer is off by a factor of 2. The correct answer is 1/12, or about 8%. Another way to see this is that there are (9 choose 3) ways to pick 3 parking spaces out of 9. That's 9!/3!6! = 84 ways. As Haversian pointed out, 7 of these ways have all three spaces together, so the probability is 7/84 = 1/12.```