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Let $A_1, A_2, A_3, ..., A_n (n>1)$ be independent events such that $P(A_i)=\frac{1}{i}$. What is the smallest possible size of $\Omega$? And what if $P(A_i)=\frac{1}{e^i}$?

I imagine this as tossing $n$ unfair coins, where the first one always lands heads and the others can land either heads or tails, hence there would be $2^{n-1}$ possible outcomes and that is the size of $\Omega$.
If $P(A_i)=\frac{1}{e^i}$, then the situation is the same, but the first coin can also land either heads or tails, hence the answer is $2^n$.

Is this correct? A similar question was asked here, but in my case one event has a probability of $1$.

Michał
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Yes. The answer you refer to shows that for $k$ independent events each with probability in $(0,1)$, the minimal size of $\Omega$ is $2^k$. If you need to add independent events with probability 0, just add $\emptyset$, and for events with probability 1, take $\Omega$ itself.

So here, if $\mathbb P(A_i)=\frac 1i$ then $\mathbb P(A_1)=1$ so the minimal size is $2^{n-1}$, and if $\mathbb P(A_i)=\frac{1}{e^i}$ then the minimal size if $2^n$.

Will
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