Answer to Question #210876 in Statistics and Probability for Nene

Question

A cane of length P is to be broken into two parts. What is the probability that one part will have a length of more than double the other? State clearly what assumptions would you have made. Discuss whether you believe these assumptions are realistic and how you might improve them if they are not.

Answer 1

Solution:

The cane has a length of P. Let t be the length of first part. 

Assumption :

t is uniformly distributed on 0 to P. This is a fair assumption as you randomly choose a point on the cane and break.

So, t ~ U(0,P)

PDF of t ,

f(t) = $\dfrac1P$  , 0<t<P

If one part is twice larger than another then two things may happen. First part is greater than twice of second part , Second part is greater than twice of first part. Now, the length of second part = P-t

Therefore we need to find the probability,

Pr({t>2(P-t)} U {P-t>2t})

=Pr[t>2(P-t)] + Pr[P-t>2t] [ as two events are mutually exclusive]

=Pr(3t>2P) + Pr(3t<P)

=Pr(t>2p/3) + Pr(t<P/3)

Now,

$\left(t<\frac{P}{3}\right) \\\quad=\int_{0}^{\frac{P}{3}} \frac{1}{P} d t \\\quad=\frac{1}{P} \times \frac{p}{3}=\frac{1}{3}$

And,

$\operatorname{Pr}\left(t>\frac{2 P}{3}\right) \\=\int_{\frac{2 P}{3}}^{P} f(t) d t \\=\int_{\frac{2 p}{3}}^{P} \frac{1}{P} d t \\=\frac{1}{P}\left[P-\frac{2 P}{3}\right]=\frac{1}{3}$

Hence probability that one part is twice greater than another part.

$\operatorname{Pr}(t>2 p / 3)+\operatorname{Pr}(t<P / 3)$

$=1 / 3+1 / 3=2 / 3$