If AB=2AC=3AE=4DB, and point x is randomly selected on line AB, what is the probabilty that x is btwn E and C?

Question

ran out of characters, btwn=between and probability spelled without a second i.
Give your answer as a common fraction.

Answer 1

I assume the points are:
A - - - - E - - C - - - D - - - B

AC = 1/2 (of AB)
AE = 1/3 (of AB)
DB = 1/4 (of AB)
So CE = 1/2 - 1/3 = 3/6 - 2/6 = 1/6

Given that all points have equal probability and the segment CE is 1/6 of the full segment (AB):

The probability is 1/6.

Answer 2

AB=2AC=3AE=4DB
means
AC = ½AB
AE = ⅓AB
DB = ¼AB

So the points in sequence must be
A ... E .. C .. D . B

AC = ½ = 6/12 = CB
AE = ⅓ = 4/12 and
DB = ¼ = 3/12
So CD = 3/12 = ¼ (because CB = 6/12 and DB = 3/12)
and EC = 2/12 = 1/6 (because AC = 6/12 and AE = 4/12)

Therefore P(EC) = 1/6

And the interval looks like this
A ....... E . C .... D .... B
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