$foundations in probability$

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$foundations in probability$

check pic please

Probability

Burgoos

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Matchmaticians foundations in probability File #1

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Burgoos

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the deadline for this is an accident, theres no 1 day
Burgoos

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can you check the other questions too?
Schwartstack

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I checked the other ones. They would take longer to answer and I don't have the time right now
Burgoos

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ok no worries thank you for your time!
Burgoos

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schwartstack can you check 5, it looks easy?

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Burgoos

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the deadline for this is an accident, theres no 1 day
Schwartstack

0

I checked the other ones. They would take longer to answer and I don't have the time right now

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Let's let X = the position of the first cut. The distribution of X is uniform(0,1).

Let Y = the position of the second cut. The distribution of Y given x is uniform(0,x). Therefore the conditional density of Y given x is 1/x, for y between 0 and x.

$E(Y) = \int_{0}^{1} \int_{0}^{x} \frac{y}{x} dy dx $

$= \int_{0}^{1} \frac{x}{2} dx$

$= \frac{1}{4}$

Schwartstack

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$foundations in probability$

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