Probability Of Three Sedans Park Adjacently To Each Other

[Zidanne]

Hello, I am new to this thread.

I have a question which asks:" There are 30 vehicles in total, 17 Sedans and 13 SUVs. If three teachers that have a Sedan each take students on a trip, then what is the probability of that three sedans will park adjacently to each other?" What I first did to solve this was 30 choose 3 because it represents the total ways 30 vehicles can be lined up in 3 parking spaces each, which is our denominator. Next I did 17 choose 3 because that's the total number of ways a Sedan can be grouped into three parking spaces, which is our numerator. When we solve this we get 34/203. I am not entirely sure if this is the answer, but if I am wrong I would like to have an explanation Thanks

 

[Dr.Peterson]

Why 17 choose 3? That might mean several things (such as which of the sedans are the teachers'), but I don't see that it gives the number of ways to park all 30 cars so that the given 3 cars are together. Can you explain your thinking?

I'd imagine first choosing a set of 3 adjacent spots (how many ways?), and then choosing the order in which those 3 cars, and all the rest, can be put in.

 

[pka]

I have a question which asks:" There are 30 vehicles in total, 17 Sedans and 13 SUVs. If three teachers that have a Sedan each take students on a trip, then what is the probability of that three sedans will park adjacently to each other?" What I first did to solve this was 30 choose 3 because it represents the total ways 30 vehicles can be lined up in 3 parking spaces each, which is our denominator. Next I did 17 choose 3 because that's the total number of ways a Sedan can be grouped into three parking spaces, which is our numerator. When we solve this we get 34/203. I am not entirely sure if this is the answer, but if I am wrong I would like to have an explanation.

Is this a joke? What does three sedans with a teacher & student on a trip have to do with parking? Nothing is given about leaving and then returning. It is quite normal in large schools to have assigned parking for faculty & staff. So what do we know about the randomness of parking?
This question is so poorly written as to be meaningless.
Please repost with more clarifying information,

 

[Zidanne]

My apologies for the badly written question, I was writing down from what I remember on this sheet of paper I lost. Let me reword the question. If there's 30 parking spaces, and there's 17 sedans and 13 SUVs, what is the probability that three sedans are adjacently next to each other? I believe it is 34/203 but I am not too sure.

 

[Zidanne]

Why 17 choose 3? That might mean several things (such as which of the sedans are the teachers'), but I don't see that it gives the number of ways to park all 30 cars so that the given 3 cars are together. Can you explain your thinking?

I'd imagine first choosing a set of 3 adjacent spots (how many ways?), and then choosing the order in which those 3 cars, and all the rest, can be put in.

My thinking was that of the 30 cars, how many of can be grouped in groups of three, then that would be the denominator since its asking what's the probability. Next the 17 choose 3 comes from how many sedans can be grouped in groups of three and so that would be the over the denominator.

Also can you elaborate(or show what you mean mathematically) of what you said?

 

[Dr.Peterson]

My apologies for the badly written question, I was writing down from what I remember on this sheet of paper I lost. Let me reword the question. If there's 30 parking spaces, and there's 17 sedans and 13 SUVs, what is the probability that three sedans are adjacently next to each other? I believe it is 34/203 but I am not too sure.

But is the question about any 3 sedans, or 3 particular sedans? That detail (the mention of the teachers) is important. And ultimately, if you don't remember the actual wording, anything we say will not be relevant to the actual problem. I don't think it's worth discussing in that case. There are just too many possibilities for what the problem says.

Your numerator and denominator both greatly oversimplify the problem. Each possible outcome has to take into account where every single car is, as well as (in the numerator) which three are being considered -- not just which three are together. If it can be any three sedans, then there might be many different sets of three together, making this a very complicated problem.

 

[Zidanne]

I forgot to add that they must occupy the first three spots of the parking lot
Sorry ?
So now the question is that there are 30 vehicles, 17 sedan and 13 SUV, what is the probability that three sedans fill the first three spots of a parking lot given that there are 30 parking spaces

 

[Zidanne]

But is the question about any 3 sedans, or 3 particular sedans? That detail (the mention of the teachers) is important. And ultimately, if you don't remember the actual wording, anything we say will not be relevant to the actual problem. I don't think it's worth discussing in that case. There are just too many possibilities for what the problem says.

Your numerator and denominator both greatly oversimplify the problem. Each possible outcome has to take into account where every single car is, as well as (in the numerator) which three are being considered -- not just which three are together. If it can be any three sedans, then there might be many different sets of three together, making this a very complicated problem.

I think this means 3 particular sedans cause it asks that three sedans must occupy the first three spots of the parking lot. Read my last comment that I posted before this

 

[pka]

I forgot to add that they must occupy the first three spots of the parking lot. Sorry
So now the question is that there are 30 vehicles, 17 sedan and 13 SUV, what is the probability that three sedans fill the first three spots of a parking lot given that there are 30 parking spaces

You need to learn to construct models. Suppose that we have seventeen green beads and thirteen red beads,
How many ways can this collection be arranged in a row that begins with three green beads.
There are \(\displaystyle \frac{30!}{(17!)(13!)}\) ways to arrange this collection of beads in a row.
Of those \(\displaystyle \frac{27!}{(14!)(13!)}\) have three green beads in the first three positions in the row.
So what is the probability that happens.

 

[Zidanne]

You need to learn to construct models. Suppose that we have seventeen green beads and thirteen red beads,
How many ways can this collection be arranged in a row that begins with three green beads.
There are \(\displaystyle \frac{30!}{(17!)(13!)}\) ways to arrange this collection of beads in a row.
Of those \(\displaystyle \frac{27!}{(14!)(13!)}\) have three green beads in the first three positions in the row.
So what is the probability that happens.

20058300?

 

[pka]

You need to learn to construct models. Suppose that we have seventeen green beads and thirteen red beads,
How many ways can this collection be arranged in a row that begins with three green beads.
There are \(\displaystyle \frac{30!}{(17!)(13!)}\) ways to arrange this collection of beads in a row.
Of those \(\displaystyle \frac{27!}{(14!)(13!)}\) have three green beads in the first three positions in the row.
So what is the probability that happens.

The answer is SEE HERE.

 

[Zidanne]

So my original answer to the original question was correct? Which was 34/203

 

[pka]

So my original answer to the original question was correct? Which was 34/203

That is the same number you posted in the OP. That number was found in counting coloured beads. You asked for an explanation. I have given you on by way of a different model.