Answer:
The probability that the longer part is at least 3 times the length of the shorter part is 0.5
Step-by-step 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)
= [tex]\int\limits^\frac{3}{4L} _\frac{L}{4} f{\frac{1}{L} } \, dx[/tex]
= 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
