Probability Using Graph

How can I solve this problem using a graph and geometric solutions?

Line segment AB has a measurement of 1. Line segments CD and EF has lengths of 0.25 and 0.1 respectively. If the two latter line segments were randomly plotted on the line segment AB (such that no lengths go off of the segment), what's the chance that they share a length?

Can someone elaborate how to do this? I'm hoping the solution doesn't involve any calculus.

#Combinatorics

Easy Math Editor

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Comments

Let $AB$ run from $x = 0$ to $x = 1$ . Let $x_1$ be the left end of $CD$ , and let $x_2$ be the left end of $EF$ . Since neither strip is allowed to extend outside of $AB$ , the range on $x_1$ is $(0 \leq x_1 \leq 0.75$ ) and the range on $x_2$ is $(0 \leq x_2 \leq 0.9$ ). This gives us a "parameter space" which is a rectangle. Divide the parameter space into a bunch of little infinitesimal squares of area $dx_1 dx_2$ , and conditionally integrate the area in parameter space.

Since $CD$ is longer than $EF$ , if they are to overlap, either the left end of $CD$ can be within $EF$ , or the right end of $CD$ can be within $EF$ . Both conditions cannot be true simultaneously. It is also possible for $EF$ to lie entirely within $CD$ . Divide the accumulated area by the total area of the parameter space to get the probability.

This can be summarized mathematically as follows:

$P = \frac{1}{0.75} \frac{1}{0.9} \int_0^{0.9} \int_0^{0.75} M(x_1, x_2) \, dx_1 \, dx_2 \\ M(x_1, x_2) = 1 \,\,\, \text{if} \,\, (x_1 > x_2) \text{AND} (x_1 < x_2 + 0.1) \\ M(x_1, x_2) = 1 \,\,\, \text{if} \,\, (x_1 + 0.25 > x_2) \text{AND} (x_1 + 0.25 < x_2 + 0.1) \\ M(x_1, x_2) = 1 \,\,\, \text{if} \,\, (x_1 < x_2) \text{AND} (x_1 + 0.25 > x_2 + 0.1) \\ M(x_1, x_2) = 0 \,\,\, \text{otherwise}$

Computing this integral results in a probability of $\approx 0.374$ . I have attached the code I used to do the computation.

import math

Num = 10**4

dx1 = 0.75/Num
dx2 = 0.9/Num

dA = dx1*dx2

Aref = 0.75*0.9

method = 1

###############################

A = 0.0

x1 = 0.0

while x1 <= 0.75:

    x2 = 0.0

    while x2 <= 0.9:

        if (x1 > x2) and (x1 < x2 + 0.1):   # if left end of CD inside EF
            A = A + dA

        if (x1 + 0.25 > x2) and (x1 + 0.25 < x2 + 0.1):   # if right end of CD inside EF
            A = A + dA

        if (x1 < x2) and (x1 + 0.25 > x2 + 0.1): # if EF entirely within CD
            A = A + dA

        x2 = x2 + dx2

    x1 = x1 + dx1

###############################

P = A/Aref

print dx1
print P

#>>> 
#0.00075
#0.374022000002
#>>> ================================ RESTART ================================
#>>> 
#7.5e-05
#0.374069190164
#>>>

Steven Chase - 1 year ago

Based on this image by someone, the probability is $\approx 0.37$ . Is there something wrong with the computation? How can I solve the problem using geometric solutions?

Kaizen Cyrus - 1 year ago

Good catch. There is a third condition that I initially forgot to consider. There is overlap if any of the three conditions are true:

1) The left end of $CD$ is within $EF$
2) The right end of $CD$ is within $EF$
3) $EF$ is contained entirely within $CD$ (neither end of $CD$ is within $EF$

When I include the third condition, I get $0.374$ as the probability. If you don't like calculus and/or programming, perhaps the other guy's geometry solution will inspire you.

Steven Chase - 1 year ago

@Steven Chase – The thing is though I don't understand how the shapes were formed/determined.

Kaizen Cyrus - 1 year ago

@Kaizen Cyrus – If you shaded the parameter space according to the rules I described, you would get those shapes

Steven Chase - 1 year ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$