Set $A_{n} = \lbrace1, 2, 3, ..., n\rbrace$
Each possible subset of $A_{n}$ has a probability $p$ to be an element of set $B_{n}$.
How much is the probability of set $C_{n}$, which is the union of all elements of $B_{n}$, is $A_{n}$?
Is there any other famous equal problem in combinatorics?
My interpretation of the problem:
Set $A_{n} = \lbrace1, 2, 3, ..., n\rbrace$
Each possible subset of $A_{n}$ has a probability $p$ to be an element of set $B_{n}$.
Find probability that set $C_{n}$, which is the union of all elements of $B_{n}$, is $A_{n}$.
Solution:
For $n = 1$, $P(C_1 = A_{1}) = p$.
Randomly choose one of the 4 possible cases for $B_1$, which may have the empty set as an element.
Construct $B_{n + 1}$ by including all elements of $B_{n}$ and then deciding if the $2^{n}$ "new" subsets of $A_{n + 1}$ would be included in $B_{n + 1}$.
The new subsets all contains the new element $n + 1$. If at least one of the new subset is chosen, then $C_{n + 1} = A_{n + 1}$. Therefore, $P(C_{n + 1} = A_{n + 1} | C_n = A_n) = 1 - (1 - p)^{2^{n}}$.
By induction, $P(C_n = A_{n}) = \prod_{n} [1 - (1 - p)^{2^{n - 1}}] $