## Akshay_Budhkar Group Title satellite plz help with the probability question!!!! 3 years ago 3 years ago

1. hafizaliimran

yup

2. Akshay_Budhkar
3. Akshay_Budhkar

try it hafiz..

4. satellite73

hello

5. Akshay_Budhkar

hello.. i tried it.. no use

6. satellite73

link isn't working for me

7. Akshay_Budhkar

we can solve here

8. Akshay_Budhkar

a rod is divided in 4 parts.. probability that it is a quadrilateral?

9. satellite73

ok it is the same problem as last time yes?

10. Akshay_Budhkar

yes

11. satellite73

not sure what you mean. i meant use the second proof in the paper i sent you. not the volume one, but the line segment one.

12. satellite73

start an page 186 at the line "construction of an n-gon will be impossible of AB is longer than 1/2" do you see where i mean?

13. satellite73

of course replace n by 4

14. satellite73

i am here.

15. satellite73

i may be easier to do this in chat, or quicker, but i can write the idea of the proof here. i am just mimicking the proof i sent you in the pfd .

16. Akshay_Budhkar

you can come in chat?

17. satellite73

you have a "rod" stick whatever and you are going to break it into 4 parts by picking 3 points uniformly at random, and the question is what is the probability that you can make a quadrilateral

18. Akshay_Budhkar

yes

19. satellite73

you already know the answer is 1/2 so far correct. so we just need the proof

20. Akshay_Budhkar

yes

21. satellite73

first trick to to say that the rod has length one so we can dispense with ratios

22. Akshay_Budhkar

ya

23. satellite73

so it looks something like this 0 --------------a-----------------b--------------------------c-------------------------1 where a, b, c are the points chosen at random to break up the rod

24. satellite73

now the question is "when would you not get a quadrilateral\"

25. satellite73

now i am just reading along what i sent, replacing n by 4

26. satellite73

construction of a quadtrilateral will be impossible if $\overline{0a}$ is bigger than one half, i.e. if all three points are to the right of the midpoint of the segment. the probability that one is to the right is 1/2 and all three to the right would be $(\frac{1}{2})^3$

27. satellite73

the same is true for the other three segments, and the probability that each of them is bigger than one half is also 1/8 so the probability of failure is 4 times 1/8 = 1/2 and therefore so is the probability of success.