# finite probability question

• Sep 9th 2009, 04:59 PM
lauren2988
finite probability question
A point X is chosen at random on a line sement AB. (a)show that the probability that the ratio of lengths AX/BX is smaller than a is = to a/(1+a). and (b) show that the probability that the ration of the length of the shorter segment to that of the longer is less than 1/3 is = to 1/2.
• Sep 9th 2009, 08:09 PM
CaptainBlack
Quote:

Originally Posted by lauren2988
A point X is chosen at random on a line sement AB. (a)show that the probability that the ratio of lengths AX/BX is smaller than a is = to a/(1+a). and (b) show that the probability that the ration of the length of the shorter segment to that of the longer is less than 1/3 is = to 1/2.

You may as well assume that the line segment is of length $1$, and that you choose a point $x\sim U(0,1)$. Now the first part of the question becomes:

find $p(x/(1-x) give $x\sim U(0,1)$.

You do this by rearranging the condition $x/(1-x) into the form $x< ...$ where $...$ is an expression in $a$ only.

CB
• Sep 9th 2009, 08:12 PM
CaptainBlack
Quote:

Originally Posted by lauren2988
A point X is chosen at random on a line sement AB. (a)show that the probability that the ratio of lengths AX/BX is smaller than a is = to a/(1+a). and (b) show that the probability that the ration of the length of the shorter segment to that of the longer is less than 1/3 is = to 1/2.

To do the second part we are being asked (same notation as before) to compute the probability that $x<1/4$ or $x>3/4$

CB