let the generator return a number, r : rєø

now consider:

A:"what's the probability of 0<r<1"

I answer: " P(A)=1"

B:"what's the probability of 0<r<=1/2

I answer:"P(B)=1/2"

We can see this as a "geometrical probability" problem. A bigger domain implies greater probability.

so lets consider the first number generated: g1

P(g1;"less than 1/4")=1/4

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P(gN," less than k")=k ; 0<k<1

But P(gN,"greater than b")=1-P(gN,"less than b)

Its a random number gen. so each gN is independent. You want first 5 to be < 1/4

P(g1;"less than 1/4") * ... * P(g5;"less than 1/4") = (1/4)^5 = 1/1024

The last 5 must be >1/2

P(g6,"greater than 1/2) * ... * P(g10,"greater than 1/2) =

= [1-P(g6,"less than 1/2)] * ... * [1-P(g10,"less than 1/2)] =

= [1-1/2]^5 = (1/2)^5 =1/32

We multiply the two results (P1 and P2)

1/1024 * 1/32 = 1/32768

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This is just an idea.

I'm stuck on what to do this kind of problem. I know how to find the probability of a number greater thanx or less than x, but I'm lost with how to work with a problem that involves the sequence.

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