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I have a 1-meter long stick. Let $x,y$ be i.i.d with uniform ([0,1]), representing the cut in the stick. What is the probability that the 3 segment form a triangle?

Attempt:

Pr{triangle} = Pr(triangle | x < y) Pr(x < y) + Pr(triangle | x > y) Pr( x > y). enter image description here

Let's look at the first term. Pr(triangle | x < y) requires:

1) x + y-x > 1-y

2) x + 1 -y > y-x

3) y-x + 1-y > x

Which is equivalent as

1.y >0.5,

2. x < 0.5

3. y - x < 0.5.

This is a triangle which has the area of 1/8.

Due to symmetry, Pr(triangle | x > y) = 1/8.

Now. Pr{triangle} = Pr(triangle | x < y) Pr(x < y) + Pr(triangle | x > y) Pr( x > y)

= 1/8 *0.5 + 1/8 *0.5 = 1/8 $\neq$ 1/4 (as posted here)

Confusion: Did I do something wrong?

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Formally, a triangle can be formed iff no piece exceeds 1/2 in length. This is correctly summarized by your equivalence.

Your calculation of the probability of this event is slightly off, since you forgot to assume that x>y is given. The triangle has area 1/8, unconditional on $x>y$. Notice that your equivalence is a subset of the event $x<y$. So:

$P(y>0.5,x<0.5,y-x<0.5|x<y)=\frac{P(y>0.5,x<0.5,y-x<0.5 , x<y)}{P(x<y)}=\frac{1/8}{1/2}=1/4.$

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  • $\begingroup$ exactly, forgot to consider P(x < y). $\endgroup$ – wrek Apr 18 '17 at 18:53

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