0
$\begingroup$

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?

Thanks

  • 0
    @H$a$ven 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.2012-12-24

1 Answers 1

0

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$