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If I have an array of $n$ numbers in the range [0,1) and have the following events for $y$:
for $i=1$ to $n$ :
generate a new number $y$ in [0,1) uniformly and independent of previous
if $y\ge0.5$ then
toss a fair coin if the coin turns up "HEADS" then
$y=y-0.5$

I need to find out what is the probability that y is in the range [0,1/n)
My solution: $1/2+1/2\cdot1/2=3/4$ - the first 1/2 is the probability when y is not greater than 0.5. The second $1/2\cdot1/2$ is the probability when $y>0.5$ AND because of the fair coin - we have 1/2 change of being HEADS.

Is it the correct way ? Do I need to add n into the equation? Please correct me if I am wrong. Thanks!

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    Your edit hasn't resolved any of my questions. You're now generating $n$ numbers $y$ instead of just one, but their relationship to the numbers in the array is still unclear.2011-11-04

1 Answers 1

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Assume $n$ is fixed beforehand. Then for each $y$ probability that it will be in range [$0,1/n$) is
$\frac{3}{4}*(\frac{1/n}{1/2})=\frac{3}{4}*\frac{2}{n}=\frac{3}{2n}$

because with probability $\frac{3}{4}$ $y$ is a random number from [$0,1/2$) and there with probability $\frac{1/n}{1/2}$ it belongs to the range [$0,1/n$).

Hereby:
1. Probability that all $y \in$ [$0,1/n$) is equal to $(\frac{3}{2n})^n$ 2. Probability that exists $y \in$ [$0,1/n$) is equal to:$1-(1-\frac{3}{2n})^n$ which is $1-$ probability that all $y \in$ [$1/n,1$)

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    What does [1/2,$n$) mean ? isn't n >= 1 always ? if so, then it's equal to probability of y being in [1/$2$,1) because y cannot be greater than 1, and that probability is 1-3/$4$=1/4.2011-11-05