satellite plz help with the probability question!!!!

- Akshay_Budhkar

satellite plz help with the probability question!!!!

- schrodinger

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- anonymous

yup

- Akshay_Budhkar

try it hafiz..

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## More answers

- anonymous

hello

- Akshay_Budhkar

hello.. i tried it.. no use

- anonymous

link isn't working for me

- Akshay_Budhkar

we can solve here

- Akshay_Budhkar

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

- anonymous

ok it is the same problem as last time yes?

- Akshay_Budhkar

yes

- anonymous

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.

- anonymous

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?

- anonymous

of course replace n by 4

- anonymous

i am here.

- anonymous

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 .

- Akshay_Budhkar

you can come in chat?

- anonymous

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

- Akshay_Budhkar

yes

- anonymous

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

- Akshay_Budhkar

yes

- anonymous

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

- Akshay_Budhkar

ya

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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
\[(\frac{1}{2})^3\]

- anonymous

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.

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