On a real line R points a,b are randomly selected such that -2<=a<=2 and 0<=b<=3. Find the probability that | a - b | > 1

Question

Let's say that C is the set where |a-b|>1

So I suppose you could say plot it as coordinates where the x-axis (labelled a) is from [-2,2] and the y-axis (labelled b) is from [0,3].

Now |a-b| must be greater than 1

|2-3| = 1

|2-2| = 0

|2-1| = 1

...

these are not in the set C

|2-0| = 2

|1-3| = 2

|0-3| = 3

|0-2| = 2

|-1-3| = 4

|-1-2| = 3

|-1-1| = 2

|-2-3| = 5

|-2-2| = 4

|-2-1| = 3

|-2-0| = 2

C = { (2,0) ; (1,3) ; (0,3) ; (0,2) ; (-1,3); (-1,2); (-1,1) ; (-2,3) ; (-2, 2) ; (-2,1); (-2,0) }

n(C) = 11

n(S) = 4 x 5

 = 20

P(C) = n(C) / n(S)

 = 11/20

 = 0.55

Is this right. Is there an easier way to do this?

Answer 1

You're on the right track: draw a rectangle with $-2<a<2 ; 0<b<3$ , on $a-b$- axes and draw the lines $a-b$=1 (associated to $x-y>1$ , and then $b-a=1$ associated to $y-x>-1$, and look at the region/subset of the rectangle where the conditions are satisfied. Calculate the area of this region and then divide by the total area of the rectangle.