Please help me with solving this and please show work. this isn't a homework assignment?

a ten foot rope is chopped at an arbitrary spot into two sections. what is the probabaility that both pieces are at least four feet long?Please help me with solving this and please show work. this isn't a homework assignment?
the bits can get cut in the following combinations (5,5);(6,4);(7,3);(8,2);(9,1), considering that the bits would be cut into exact feet and not fractions of it.





so we have 5 combinations, of which there are only 2 combinations where we can have both the bits measuring atleast 4.


therefore the chances are 2 by 5 ,


which is .4 or 40%Please help me with solving this and please show work. this isn't a homework assignment?
o/10 becuz the answer is 5/


5
100% BOTH HAS TO BE FIVE FOOT.
well it could go either way it depends on what the length of the first one is to see if the other is longer than 4 feet.





Example:


If the first one is 5 feet, that makes the other five feet, so yes it would by more than 4 feet. But if the first one is say 7 feet, then the 2nd one is going to be 3 feet so it would not be longer than 4 feet. Do you understand it better now?
Everyone before talks about cutting the rope in a real finite spot, which works in the real world. However, you said choose an arbitrary point. By that logic you could cut the rote at an infinite number of locations, giving you and infinite amount of cuts that allow both pieces to be at least 4ft and an infinite amount of cuts that allow the rote to be in pieces that do not satisfy the condition.
There is a 40% probability that one of the sections will be 4 feet or less. This can occur on either side of the rope. So 100% - 2*40% = 20% that both pieces are four feet or greater.
The arbitrary spot should lie between 4 and 5 feet or 5 and 6 feet.


0-1,1-2,2-3,3-4,4-5,5-6,6-7,7-8,8-9,9-鈥?br>

2/10 probability.


2/10=1/5=0.2