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when 3 fair dices roll, let Y be a Random Variable of which value is the minimum result of all 3 dices results. for example if the results are (2,3,2) then Y=2.

how do i find, in the most elegant and reasonable way the P(Y) function?

i tried tackle this for a while, so the more explanations the better! thanks!

this is my professor answer (which i don't get).

(ignore the related foreign letters)

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up vote 2 down vote accepted

He is calculating the cumulative distribution first. That is, he is finding, for instance, the probability that the minimum die roll is greater than or equal to 2. This is an easier quantity to calculate: the minimum die roll is greater than or equal to 2 if and only if all three die rolls were greater than or equal to 2. So, the probability that the minimum die roll is greater than or equal to 2 is $p(Y\ge 2)=({5\over6})^3$ (we know each die roll was 2 or greater. The rolls are presumed independent and the probability that a particular die is roll is equal to 2 or greater is $5/6$).

Then the probability that the minimum die roll is equal to $i$ is calculated using the cumulative distribution. So, for instance $p(Y=1)=p(Y\ge 1)-p(Y\ge 2)$.

The last formula can be thought of as follows: take the set of outcomes $A=[Y\ge 1]$. Then this set of outcomes differs from the set of outcomes $B=[Y=1]$ by exactly the set of outcomes $C=P[Y\ge 2]$. That is, $A=B\cup C$ and $B\cap C=\emptyset$. We thus have $P(A)=P(B)+P(C)$. Rearranging this gives $P(B)=P(A)-P(C)$.

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that is a well explained answer, thanks a lot! could you reffer to my edit too? – Ofek Ron Mar 4 '12 at 22:45
A=B∪C and B∩C=∅ what sentence\property were you laying on for this to be true? – Ofek Ron Mar 4 '12 at 22:51
@OfekRon It should be fairly clear from the definitions of $A$, $B$, and $C$, that $A=B\cup C$ and $B\cap C=\emptyset$. Take the set where $Y\ge 1$, throw away all the elements with $Y\ge 2$. The elements that remain are those with $Y=1$. – David Mitra Mar 4 '12 at 22:58

$P(X=1)=P(X\leq1)=1-P(X >1)=1-P(X\geq2)$ this means that the possibility of X=1 is the possibility of not being greater than 2. Due to the fact that the 3 throws are independent $P(X\geq2)=\prod\limits_{i=1}^3 P_i(X\geq2)=(\frac{5}{6})^3$ because 5 out of 6 throws a dice will roll a number greater or equal to 2. So you got:$P(X=1)=1-(\frac{5}{6})^3=\frac{91}{216}$. That's the principle for the other calculations too.

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it should be multipication instead of a summation is it not? – Ofek Ron Mar 4 '12 at 22:32
Yeah,sorry it was a typo.I corrected it. – chemeng Mar 4 '12 at 22:35
plz reffer to my edit... – Ofek Ron Mar 4 '12 at 22:35
I didn't quite understand your formula but I think the Inclusion-Exclusion Principle states: $$|A\cup B\cup C|=|A|+|B|+|C|-|A\cap B|-|B\cap C|-|A\cap C|+|A\cap B\cap C|$$ So we got $P(X=1)=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}-\frac{1}{6}\cdot\frac{1}{6}-\frac{1}‌​{6}\cdot\frac{1}{6}-\frac{1}{6}\cdot\frac{1}{6}+\frac{1}{6}\cdot\frac{1}{6}\cdot\‌​frac{1}{6}=\frac{91}{216}$ – chemeng Mar 4 '12 at 22:53
isnt X~B(3,1/6)? – Ofek Ron Mar 4 '12 at 22:53

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