Hi! I am a new member, and I'm posting b/c I have a question from the practice list I have received as a practice for an upcoming quiz.(adsbygoogle = window.adsbygoogle || []).push({});

I've done many questions where mass x (or y) is dependent of x. However, I've come across one where the length of a piece x (of two) is completely independent of x:

Suppose you have a stick of length L that you would like to cut into two pieces. Assume that the cutting is completely random, that the (1)probability density p(x)for having either one of the pieces be of mass x is completely independent of x.

Find the function p(x).

So, I did integral (from x =0 to x=L) p(x)dx = integral (x=0 to x = L) C dx = 1 for normalization. (I put C as a constant).

Then, I equivelated 1 = C integral (x=0 to x = L) dx = C x/(x=0 to x = L)

which becomes:

1 = C (L - 0) = CL

C = 1/L

...and, from this, how am I to get thefunction p(x)? Do I just put p(x) = 1/L?

Also, in case I want to find the(2) average length of one piece(preferably a? w/out L/2 value), how do I solve it? For this, I did the similar thing as above, except:larger one

integral (from y = L/2 to L) p(y) dy = integral (y = L/2 to L) C_new dy = 1 for normalization (I put y = larger piece)

THen I equivalted 1 = C_new integral (y = L/2 to L) dy

1 = C_new (L - L/2)

1 = C_new (L/2)

C_new = 2/L....

So, I wonder if average is 2/L, from 2/L integral (y=L/2 to L) y*dy? Or, is it L/2?

Any suggestions or replies would be welcome!

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# Probability Question_cutting one into two .

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