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a. A point is chosen uniformly at random on a line segment of length L, dividing the segment into two parts. Find the probability that the longer of the two parts is at least twice as long as the shorter part.

b. Again a point is chosen at random on a line segment of length L, dividing the segment into two parts. This time the location of the random breaking point has pdf:

$$f_X(x) = { 6x(L-x) \over L^3}, where: 0 \le x \le L $$

Find the probability that the longer of the two parts is at least twice as long as the shorter part.

What I have tried so far:

a. I am really stuck on where to begin. Could someone give me a start in the right direction.

b. same as part (a)

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2 Answers 2

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Hint:

a) Actually you are asked to find $P(X\leq\frac13L\vee X\geq\frac23L)$ for $X$ having uniform distribution on $[0,L]$.

b) Similar situation with other distribution.

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a. You are looking for $Pr\{X/(L-X)\geq 2\}+Pr{(L-X)/X)\geq 2}$; after transformation, it is exactly $Pr\{X\leq 1/3L\}+Pr\{X\geq 2/3L\}$, which can be obtained easily through the CDF.

b. Similar to a., you need CDF of the distribution. It is easy since you already know the PDF of $X$

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