Question

Divide a line segment into two parts by uniformly selecting a point inside the segment. Determine the probability that the longer part is at least 3 times the length of the shorter part. Hint: Use interval [0,L] to denote the line segment. Let X be the dividing point. How to represent the event {the longer part is at least 3 times the length of the shorter part} in terms of X.

Answer 1

The probability that the longer part is at least 3 times the length of the shorter part is 0.5

Explanation:

Let us assume,

X be randomly selected the dividing point in the interval [0,L]. So, X has a uniform distribution with pdf

f(X) = 1/L ( where X is greater or equal than 0 and L is greater or equal than L)

The length of the complete line segment is L and length of the segment to the left of X is X and to the right is L-X

There are 2 cases ,

- The left segment is shorter than the right one. Then the length of right segment (which is L-X) should be at least 3 times the length of the left segment (which is X). So,

L- X < or = to 3X

and then , X < or = L/4

- The left segment is the longer one. Then the length of left segment (which is X) should be at least 3 times the length of the right segment (which is L-X). That is,

3 (L- X ) < or = to X

and then , X > or = L *3/4

We can say that the longer part is at least 3 times the length of the shorter part if

X > or = L *3/4 or X < or = L/4

The probability that the longer part is at least 3 times the length of the shorter part is

P(X > or = L *3/4 U X < or = L/4) = 1 - (X > or = L *3/4 U X < or = L/4)

=
`\int\limits^(3)/(4L) _(L)/(4) f{(1)/(L) } \, dx`

= 1 - 1/L (3/4L - L/4)

= 1 - 0.5

=0.5

The probability that the longer part is at least 3 times the length of the shorter part is 0.5