Two insurers provide bids on an insurance policy to a large company. The bids must be between 2000 and 2200 . The company decides to accept the lower bid if the two bids differ by 20 or more. Otherwise, the company will consider the two bids further. Assume that the two bids are independent and are both uniformly distributed on the interval from 2000 to 2200. Determine the probability that the company considers the two bids further. This is a question from an Actuary Exam. (Answer $0.19$)
My solution is the following:
$$\begin{align*}P(|X-Y|<20)&=P(X-20<Y<X+20)\\[10pt] &=\int_{2000}^{2020}\int_{2000}^{x+20}\frac{1}{200} \, dy \, dx + \int_{2020}^{2180}\int_{x-20}^{x+20}\frac{1}{200} \, dy \, dx + \int_{2200}^{2180}\int_{x-20}^{2200}\frac{1}{200} \, dy \, dx \\[10pt] &=3+32+3=38\end{align*}$$ when I saw my answer I immediately knew it was wrong because it is not less than 1. If I divided 38 by 200 I will get 0.19 as desired but why do I need to divide by 200 if $$p(x)=\begin{cases} \frac{1}{b-a} & x\in[a,b]\\ 0 & \text{otherwise} \end{cases}$$ because $f_{XY}((x,y)=f_x(x)f_y(y)$ sorry about the first part I realized the answer after I posted.
Then I thought what I am really doing is finding the cdf so why not use that. So then I got the following $$\begin{align*}P(X-20<Y<X+20)&=P(Y<X+20)-P(Y<X-20)\\&=\frac{x+20-2000}{200}-\frac{x-20-2000}{200}\\&=\frac{40}{200}\end{align*}$$ now this one is off by $\frac{2}{200}$ so my question is why is my first attempt off by 200 and why is the second attemp off as well?
The second method of solution that you wrote does not work, because $$\Pr[Y < X + 20] \ne \frac{X + 20 - 2000}{200}$$ if $2200 \ge X > 2200 - 20$, since you would get a probability exceeding $1$. Similarly, $$\Pr[Y < X - 20] \ne \frac{x - 20 - 2000}{200}$$ if $2000 \le X < 2000 + 20$, as this would give you a probability less than $0$. Instead, you must take into account these cases to get an accurate result.
It is perhaps easier to observe that $$\Pr[|X - Y| < 20] = 1 - \Pr[|X-Y| \ge 20],$$ and because $X, Y \in [2000, 2200]$, the condition $|X-Y| \ge 20$ implies that either $$2200 \ge Y > X + 20 \ge 2020$$ or $$2200 \ge X > Y + 20 \ge 2020,$$ and that these are symmetric about $Y = X$, it is easy to compute the probability of one of these events, multiply by $2$, and then compute the complement.
Perhaps another way of understanding the question is to sketch the support of $X$ and $Y$, and then shade in the region for which the desired event occurs:
This now clearly shows how you should calculate the desired probability.