521 Math #9: Broken Stick Problem (2 approaches)

This probability problem can be solved using various methods. In this videos, 2 different approaches are provided. The first approach is by geometrical method, and the second approach is via algebra, which is visualized using Geogebra. Both method are elementary, without Calculus.

There is an extension at the end of the video. Please give it a try.

If you have other method to solve it please share with us in the comment.

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A math problem from previous week will be solved and discussed on every TUESDAY. Thereafter another math problem will be posted and to be discussed the following Tuesday. These problems are suitable for math enthusiasts, it helps in learning and sharpening skills in math Olympiad skill too.

Each video is about 2-3 minutes, no talking, just some music. Check it out and share to anyone who may like it. Please comment if you have any idea to improve the videos, or any alternative solutions to the problems.

Have fun and enjoy the problem solving!

[Fpscmath]

Let length of stick is 10m We divide it into 3 pieces then the combination of lengths of pieces whose sum is 10 will be

811, 721, 622, 631, 541, 532, 442, 433

(ignoring there positions le. 811-118=181)
this is sample space

n(S) = 8

Now according to conditions of rectangle no single piece length must exceed 5m and the sum of length any two pieces will be greater then 5.
Only two combinations fulfill the condition and they are 442, 433

So n(A) = 2

P(A) = n(A)/P(S)

P(A) = 2/8
P(A) = 1/4

[DJ-xbfk]

I wonder if 100 random guesses would give a good estimation.