Using uniform distribution to determine the side of a die

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SUMMARY

This discussion focuses on simulating the outcome of a die roll using a uniform distribution U(0, 1). The simulation involves generating a random number and mapping it to die outcomes based on specified intervals. The second part defines a new variable Y, calculated as round(6U + 1), which results in possible outcomes of 1 through 6, each with a probability of 1/6. The confusion arises regarding the possibility of Y equaling 7, which is clarified as having a probability of 0.

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  • Understanding of uniform distribution, specifically U(0, 1)
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  • Basic concepts of probability, including discrete outcomes
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Homework Statement


Let U have a U(0, 1) distribution.
a. Describe how to simulate the outcome of a roll with a die using U.
b. Define Y as follows: round 6U + 1 down to the nearest integer. What are
the possible outcomes of Y and their probabilities?


Homework Equations


A continuous random variable has a uniform distribution on the interval [α, β] if its probability density function f is given by f(x) = 0 if x is not in [α, β] and
f(x) = 1 / (β − α) for α ≤ x ≤ β.
We denote this distribution by U(α, β).

The Attempt at a Solution


I'm pretty sure I have part a. Generate a random number u, if:
u <= 1/6, the die is 1
1/6 < u <= 2/6, the die is 2
2/6 < u <= 3/6, the die is 3
3/6 < u <= 5/6, the die is 4
4/6 < u <= 5/6, the die is 5
5/6 < u, the die is 6

But I'm confused with part b. Obviously, 1, 2, 3, 4, 5, 6 and possibilities such that
P (Y = 1) = P (Y = 2) = P (Y = 3) = P (Y = 4) = P (Y = 5) = P (Y = 6) = 1/6
But isn't P(Y = 7) a possibility? But what is it's probability?
 
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Y=7 is a possibility, but P(Y=7) = 0.

For U(α, β), if E = {x1,x2,...,xn} is any subset of [α, β] such that E has only a finite number of elements, then P(E) = 0.
 

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