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

Edit: I edited the question to make it more clear. Note that apart from the clear answer to this question from mjqxxx, Gerry Myerson gives an answer to the question "If two numbers $a$ and $b$ are chosen uniformly and independently in $[0,10]$, what is the probability that the product $ab>4$". Both answers are nicely illustrated by Henry. Gilles Bonnet 17.05.14.

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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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    Also tell me why the other ans i.e area outside the curve AB=4 didn't work.2012-06-27
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To emphasize Gerry Myerson's answer

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$\int_{4/10}^{10} \left(10-\frac{4}{x}\right) \, dx \quad / \quad 10^2$ $= 0.96 - 0.04 \,\log_e 25 \approx 0.831\ldots$

And for mjqxxxx's alternative interpretation of $AB \gt 4$

enter image description here

$2 \times \frac{6 \times 6}{2}\quad / \quad 10^2 \quad = \quad 0.36$

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    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.2012-06-27
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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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    @mjqxxxx, darn, looks like you're right. OK, everyone, ignore my answer and comment.2012-06-27