probability
Two points $a$ and $b$ are randomly selected such that $-2 \leq a \leq 1$ and $0 \leq b \leq 2$.
Find the probability that the distance between $a$ and $b$ is greater than $2$.
(Answer: $\frac{1}{3}$)
I'm getting $\frac{1}{4}$
Am I wrong or is the solution wrong.
I plotted it on a system of 2 axes. One with $a$ and the other with $b$.
I drew a dotted linear line from $(-2, 0)$ to $(0,2)$. I saw 3 of the 12 points being greater than having a distance of 2. Are there more?
It is not possible to pick an integer in the range $[1,\infty)$ in such a way that each possible selection has equal probability, and you have discovered one of many arguments for this impossibility.
More specifically, assume that there was such a probability distribution. If the shared probability $p$ of choosing any particular number is greater than $0$, then it must also be greater than $\frac{1}{n}$ for some $n$ (that is how real numbers work). But then the total probability of getting one of the numbers $1$, $2$, $3,\ldots, n$ would be greater than $1$ which is absurd.
On the other hand, if $p=0$, then the probability of getting any natural number at all would be $0$ (which is also absurd). This is because probability distributions are required to be countably additive.
Therefore $p$ can neither be $0$ nor different form $0$, which is a contradiction; we must conclude that a probability distribution over the naturals where each outcome has equal probability is not possible.
Since you are integrating a constant function, it's going to be so much easier if you draw yourself a picture of the area and use simple geometry, rather than integrating an analytic expression:
And dividing the hatched area by the total square area, the resulting probability is $\frac{7}{36}/\frac{1}{4} = \frac{7}{9}$
Best Answer
As you have done, draw a rectangle (whose area is $6$) of the given domain in the $ab$-plane. The desired region we want is when $|a - b| > 2$ so that either $b > a + 2$ or $b < a - 2$. Only the first inequality is relevant here; shade in the part of the rectangle that satisfies it. You'll get a triangle whose area is $2$, which explains why the answer is: $$ \frac{2}{6} = \frac{1}{3} $$