Let "X=" the length of the shorter piece of rope, so that "X\\leq 6\\ m."
The random variable "X" is uniformly distributed on the interval "0\\leq X\\leq 6," so
If "\\dfrac{X}{12-X}\\leq\\dfrac{1}{3}," so
"X\\leq3"
"P(0\\leq X\\leq 3)=\\dfrac{3}{6}=\\dfrac{1}{2}"
The probability that the length of one piece is at least thrice the length of the other is "\\dfrac{1}{2}."
Comments
Leave a comment