Question

Uniform distribution with a = 0 and b = 1.

A stick of wood with length 1 is cut at a random place into 2 parts. What is the probability that the longer part is more than twice as long as the shorter part?

Answer 1

Let X be the place at which the wood was broken. Then X ~ Uniform(0, 1)

Let the U and V be the length of longer and shorter part. Then,

U = 1 - X for X < 1/2

= X for X $\ge$ 1/2

and

V = X for X < 1/2

= 1 - X for X $\ge$ 1/2

Probability that the longer part is more than twice as long as the shorter part = P(U > 2V)

= P(X < 1/2) * P(1-X > 2X | X < 1/2 ) + P(X $\ge$ 1/2) * P(X > 2(1-X) | X $\ge$ 1/2)

= P(X < 1/2) * P(X < 1/3 | X < 1/2) + P(X $\ge$ 1/2) * P(X > 2/3 | X $\ge$ 1/2)

= P(X < 1/2) * P(X < 1/3 and X < 1/2) / P( X < 1/2) + P(X $\ge$ 1/2) * P(X > 2/3 and X $\ge$ 1/2) / P(X $\ge$ 1/2)

= P(X < 1/2) * P(X < 1/3) / P( X < 1/2) + P(X $\ge$ 1/2) * P(X > 2/3) / P(X $\ge$ 1/2)

= P(X < 1/3) + P(X > 2/3)

= (1/3) + (1 - 2/3)

= 2/3