# Probability of selecting 2 points on a straight line

On a straight line of length 10 cm, two points A, B are selected at random. What is the probability that $AB > 4$?

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Draw Cartesian axes labeled $A$ and $B$. You're asking about the fraction of the square $[0,10]\times[0,10]$ that is greater than $4$ units away (measured horizontally or vertically) from the diagonal $A=B$. Geometrically this region consists of two isosceles right triangles with side length $6$, so its total area is $36$. The desired probability is $0.36$.

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Can you please put a diagram.It will be easy to understand.Your ans is right. –  Aizen Jun 27 '12 at 7:49
Also tell me why the other ans i.e area outside the curve AB=4 didn't work. –  Aizen Jun 27 '12 at 7:54

Draw Cartesian axes, labeled $A$ and $B$. You're asking about the part of $[0,10]\times[0,10]$ outside the curve $AB=4$.

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A quick Monte Carlo simulation tallying proportion of pairs of uniform pseudoramdom points in [0,1] separated by more than 4/10 seems to converge to ~0.36, and this is approximately the numerical value of the integral of 4/(10x) from 4/10 to 1. (ie, 2 Log(5/2)/5). but is that the same as the "part outside" the curve AB=4? –  alancalvitti Jun 27 '12 at 5:51
There's a square, there's a curve inside the square, and I want the proportion of the square above the curve. But why are you looking at numbers separated by more than 4/10, when the question is about products, not differences, of numbers? –  Gerry Myerson Jun 27 '12 at 6:45
@GerryMyerson: Because the notation "$AB$" means the length of the segment from $A$ to $B$. Admittedly, the problem should ask for the probability that $AB > 4$ cm, not just "$4$". –  mjqxxxx Jun 27 '12 at 7:31
@mjqxxxx, darn, looks like you're right. OK, everyone, ignore my answer and comment. –  Gerry Myerson Jun 27 '12 at 11:32

$$\int_{4/10}^{10} \left(10-\frac{4}{x}\right) \, dx \quad / \quad 10^2$$ $$= 0.96 - 0.04 \,\log_e 25 \approx 0.831\ldots$$
And for mjqxxxx's alternative interpretation of $AB \gt 4$
$$2 \times \frac{6 \times 6}{2}\quad / \quad 10^2 \quad = \quad 0.36$$
This does not show the points where $AB > 4$. Note that this is a distance, not the product of the coordinates of $A$ and $B$. –  mrf Jun 27 '12 at 9:31
I think where the original question says $AB$, it means the distance between points $A$ and $B$ on the segment, not the product of the distances of $A$ and $B$ from the left endpoint of the segment. –  MJD Jun 27 '12 at 13:51