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I posted a similar probability question before but I realized I missed something, so here's another question for you folks.

Say, I have a set of $10,000$ numbers which, if chosen $10,000$ times, have a probability of $\frac{1}{3.3}$. Some numbers may have a probability of $\frac{1}{4}$ some $\frac{1}{2}$ some $\frac{1}{5}$ but the total of all 10,000 numbers will be $\frac{1}{3.3}$. So if I do $30.3$ thousands draws I will win $\frac{1}{3.3}$ times.

Now say, $2500$ numbers of those $10,000$ numbers has a probability of $\frac{1}{5.3}$.

Now, in order for the other all these $10,000$ numbers to have a total probability of $\frac{1}{3.3}$, the other $75000$ must have a probability of less than $\frac{1}{3.3}$, otherwise the total wouldn't equal $\frac{1}{3.3}$.

How do I calculate the probability of the other $7,500$ numbers?


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What do you mean by "A set of 10.000 numbers have a probability of 1/3.3" ? – Applied mathematician Dec 24 '12 at 13:24
THis sentence needs clarification: "Say, I have a set of 10,000 numbers which, if chosen 10,000 times, have a probability of $\frac{1}{3.3}." What does that mean? – Thomas Andrews Dec 24 '12 at 13:24
It means that, for instance, out of 30,333 draws these 10,000 numbers (say, from 1-10,000) will show up once - ON AVERAGE. It does not mean that each one of these 10,000 numbers will show up one time in the 30,333 draws. It's possible that number "1" will show up 2 times and number "2" will show up zero times. However the average of all these 10,000 numbers will remain 3.3. So out of the 30,333 draws we will have 10,000 numbers from 1-10,000 – Mike Dec 24 '12 at 13:35
@Mike and what about the other 20,333 draws? – Hagen von Eitzen Dec 24 '12 at 13:41
@Haven Von Eitzen the other 20,333 draws will be numbers above 10,000. But I'm assuming we choose only from 1-10,000. In other words, we loose those 20k + draws. – Mike Dec 24 '12 at 13:43

The formulation is a bit unclear, but suppose we have the follwoing events $$A\equiv\text{The number drawn is among the first 2500 numbers}$$ $$B\equiv\text{The number drawn is among the other 7500 numbers}$$ Then it seems we are given the value of $P(A\cup B)=\frac 1{3.3}$ and $P(A)=\frac1{5.3}$. Then one can compute (since $A\cap B=\emptyset$) $$ P(B)=P(A\cup B)-P(A)=\frac1{3.3}-\frac1{5.3}=\frac{200}{1749}\approx 0.114$$

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