# Probability - select numbers in order

I'm trying to learn statistics from a school text book, and was presented with two questions that reads:

1. Imagine you have a list of $30$ different cars for sale - all with different buyout prices. You are allowed to select two random cars from that list without looking. What would the probability be if you selected the cheapest and the most expensive car?
2. What would the probability be if you selected the most expensive car first and the cheapest second?

If I've learned anything at all in this course, I'm thinking that the first question could be solved by doing:

$$P(A) = (1/30) = 0.033...$$

$$P(B) = (1/30) = 0.033...$$

$$P(A) * P(B) = 0.0011...$$

However, for the second question, I'm not so sure. I think that you would calculate it like this, but I'm not confident enough to be sure, and the text book doesn't give me any answers, so... here goes:

$$P(A) = (1/30) = 0.033...$$

$$P(B) = (1/29) = 0.034...$$

$$P(A) * P(B) = 0.0011...$$

I'm almost certain that I'm wrong on the second assignment, but I'm thinking that since you want the most expensive car first, you'd have a $\frac{1}{30}$ chance to pick it. When its picked, there will be $29$ cars left, leaving you with a $\frac{1}{29}$ chance to pick the cheapest car. Or have I got all this wrong?

• @turkeyhundt You should probably put that as an answer. – AlgorithmsX Dec 4 '16 at 16:47
• Deleted my comment and moving it to an answer with additional info regarding your comment. – turkeyhundt Dec 4 '16 at 16:47

So, to select the cheapest and most expensive cars you could either select the cheapest, then the most expensive ($\frac{1}{30}\times\frac{1}{29}$) OR select the most expensive then the cheapest ($\frac{1}{30}\times\frac{1}{29}$). The word "or" suggests adding these together to get your answer. $$\frac{1}{30}\times\frac{1}{29}+\frac{1}{30}\times\frac{1}{29}$$
Alternately, you can look up how many ways to choose two from 30 (combinations). This will be the general way you will want to think of these problems going forward. Only one combination of two cars will be the one with the cheapest and most expensive so your answer will be $$\frac{1}{30\choose2}$$