There are 38 cars sitting on Dandy Dan's Used Car lot. 14 are SUVs, the rest are sedans. If we select 7 cars at random, what is the probability that we select exactly 4 SUVs?

This is about combinations.

How many ways are there to pick 7 cars from 38 cars?
How many ways are there to pick exactly 4 SUVs from the 38 cars?

Divide the second by the first to get the answer. Can you post what you get? :)

Don't you have to use binomial distribution?

Nope, since as soon as you pick the first car, the probability changes, and this is one of the conditions that you must not allow to happen when using binomial distribution.

Can you show me how its done?

Do you know how many ways are there to pick 7 cars out of a total of 38?

For example, if there are 5 fruits, 2 being apples and you pick 3 random fruits. (the rest being 'non apples')

The total number of ways to pick 3 fruits is given by 5C3 = 10
Or in expanded:
3 non apples
Non apple1, Non apple2, Apple1
Non apple1, Non apple3, Apple1
Non apple2, Non apple3, Apple1
Non apple1, Non apple2, Apple2
Non apple1, Non apple3, Apple2
Non apple2, Non apple3, Apple2
Non apple1, 2 apples
Non apple2, 2 apples
Non apple3, 2 apples

Notice that order in not important here.

And if you don't know what 5C3 is, it is calculated as follows:



There are the same numbers of numerators as there are denominators.

For example:

If you understand that, you can understand the first part. Can you find that out first? :)

So it would be 38C7? Which would be 12,620,256?

Good! Now to the next part, how many ways are there to pick 7 cars, 4 of which must be SUVs.

You need to pick the 4 SUVs from the 14. This is the same as before, can you work that out?

Then, since you picked 4 cars, you still need 3 more. So, pick 3 cars from the 24 (obtained by 38 - 14) other cars.

Add both of those results to get the total results of getting exactly 4 SUVs in the 7 selected cars. Can you post what you get for each step? :)

EDIT: Mistake here, don't add, but multiply. Having one SUV in one way, gives many other different ways with other cars. Hence the multiplication.

14C4 = 1001
24C3 = 2024

1001+2024 = 3025

Is this right? Thanks so much!

Right! Now, simply divide that value by the total which you got earlier on and you'll be done with the question :p

Post what you get one last time :)

EDIT: See comment later on

0.00023969?

Great! Well done dpp34! :cool:

Thank you so much for the help! Still, I don't get why other tutors have been telling me to use binomial distribution? When do I use binomial distribution? Can you give me an example?

So, the answer to the following problem would be 0.00016335, correct?


There are 34 cars sitting on Dandy Dan's Used Car lot. 17 are SUVs, the rest are sedans. If we select 9 cars at random, what is the probability that we select exactly 4 SUVs?

My Professor says that that answer is wrong?

Okay, for example, you could have it adapted like that:

There are 38 cars coming from a parking lot. You don't know how many SUVs there are, nor how many non-SUVs, but you were told that the probability that a SUV gets out is 0.23. Out of the first 7 cars, what would be the probability of getting exactly 4 SUVs.

As you can see here, the probability that a SUV gets out at any instant is 0.23, irrespective of how many SUVs got out already (if any).

Then, it would be:
Let X be the event that a SUV gets out.
X ~ B (7, 0.23)

Then,



And this would give you the answer.

I just got from a shower, and I think I missed something... maybe I'm too tired :(

I'll recheck those.

So, you do use the binomial distribution?

No, this will not work.

Well, to be honest, combinations and permutations was never my strong point at school... but using the conventional, rather long way:



From that, you know that there are ways of arranging those, that is... this makes 35 probabilities to calculate if you worked it out.

S being a SUV and N being a non-SUV and each represent the order they would appear. You will need to calculate each probability. Okay, let's see the first.
P(S, S, S, S, N, N, N) = P(1st is SUV) x P(2nd is SUV) x... P(7th is Non-SUV)
= 14/38 x 13/37 x 12/36 x 11/35 x 24/34 x 23/33 x 22/32

And for the second, you would have P(S, S, S, N, S, N, N) = 14/38 x 13/37 x 12/36 x 24/35 x 11/34 x 23/33 x 22/32

As you can see, there are only two swapped figures, but at the end, the net probability of those remain the same. The same will happen for the others, so you have to multiply one probability by 35 to get the final answer.

Ans: (14/38 x 13/37 x 12/36 x 11/35 x 24/34 x 23/33 x 22/32) x 35 = 0.1605374... = 0.161

And I'm so sleepy right now :(

Quote:

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Originally Posted by ddp34

14C4 = 1001
24C3 = 2024

1001+2024 = 3025

Is this right? Thanks so much!

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Oh, I got it, sorry for the inconvenience. This should be 1001*2024 instead of adding them.

Then you get the correct answer. Sorry again, I'm just very tired :o

So then what would be the answer of the following:

There are 31 cars sitting on Dandy Dan's Used Car lot. 15 are SUVs, the rest are sedans. If we select 6 cars at random, what is the probability that we select exactly 2 SUVs?

I really appreciate the help. Thank you.