Problem solving: Counting and probability

Question

i am a little bad at probability, i'm studying to overcome this lack. Since i'm not with a tutor i need some help on the correct way to approach a basic probability problem. I would appreciate your help understanding how to deal with those problems. Thanks a lot.

An automobile business, receives a shipping with 20 new cars. Among them, there are 7 Kia cards, 6 Mazda cars, 4 Hyundai cars, and 3 Toyota cars. Every vehicle is different. The agency has a showroom with 4 labeled spots for exhibitions. For the manager, it is important to pick 4 cars for the exhibition.

• In how many ways you can arrange the cars for the exhibition?
• What is the probability that if we choose 4 random cars, they are from different brands?
• What is the probability that if i choose 4 random cars, 2 of them were Kia?
• What is the probability that in a random arrangement in the two first spots the cars were Kias and in the two last spots, different than Kia brands?

This is what i have advanced so far:

1) We can choose 4 out of 20 elements in $P_{4, 20}$ ways, so that is the count of possible permutations for cars.

2) Since the order of picking the elements is not important, we need to use combinations.

3) $\frac{\dbinom{7}{2}\dbinom{13}{2}}{\dbinom{20}{4}}$

4) [K K NK NK] There are 7 possible choices for the first spot, 6 for the second, 13 for the third, and 12 for the last one.

I'm approaching this correctly?. Thank you very much.

Answer 1

For question 1)

I think you mean $P_{4,20}$ That is correct.

For questions 2):

In general if you pick $c=k+m+h+t$ cars then the probability that $k$ of them are Kia, $m$ are Mazda, $h$ are Hyundai and $t$ are Toyota is:

$$\binom{7}{k}\binom{6}{m}\binom{4}{h}\binom{3}{t}\binom{20}{c}^{-1}$$

In your case $k=m=h=t=1$ and $c=4$ resulting in $7\times 6\times 4\times 3\times\binom{20}{4}^{-1}$

The same principle can be applied to answer 3) where you must split up in Kia and not-Kia, resulting in: $$\binom{7}{k}\binom{13}{c-k}\binom{20}{c}^{-1}$$

In your case $k=2$ and $c=4$

For question 4) let it be that you choose $4$ cars where the first chosen gets place $1$, the second place $2$ et cetera.

That gives probability $$\frac{7}{20}\frac{6}{19}\frac{13}{18}\frac{12}{17}$$ for getting $2$ Kia's on the first two spots and $2$ non-Kia's on the others.

Answer 2

1) We can choose 4 out of 20 elements in $P_{2, 20}$ ways, so that is the count of possible permutations for cars.

Assuming you meant ${^{20}P_{4}}$, then yes.

2) Since the order of picking the elements is not important, we need to use combinations.

Yes you do. What combinations do you plan on using?

3) $\frac{\dbinom{7}{2}\dbinom{13}{2}}{\dbinom{20}{4}}$

Yes.

4) [K K NK NK] There are 7 possible choices for the first spot, 6 for the second, 13 for the third, and 12 for the last one.

So: $\frac 7{20} \frac 6{19}\frac {13}{18}\frac{12}{17}$, yes.