Events occuring Let $A_1,....A_{20}$ be mutually independent events and let $p_i$ be there probabilities $(p_i = P(A_i))$ , $i  = 1,2,...,20$.
Express $Pr$(no events occuring) in terms of $p_1...p_{20}$
Express $Pr$(all events occurs) in terms of $p_1...p_{20}$
Express $Pr$(exactly one of the event occurs )in terms of $p_1...p_{20}$
I really got stuck on all three questions
 A: The probability that $A_i$ does not happen is $1-p_i$. 
Therefore by independence the probability that $A_1$ does not happen and $A_2$ doesn't happen and so on up to $A_{20}$ does not happen (that is, the probability that none of the $A_i$ happen) is
$$(1-p_1)(1-p_2)\cdots (1-p_{20}).$$
By the way, though you seem not to have been asked this, the probability that at least one of the $A_i$ happens is therefore $1-(1-p_1)(1-p_2)\cdots(1-p_{20})$. 
The question about the probability all the $A_i$ happen is the simplest. The probability is, by independence, the product $p_1p_2\cdots p_{20}$. 
For the probability of exactly one, note that the probability that $A_1$ happens and the others don't is $p_1(1-p_2)(1-p_3)\cdots(1-p_{20})$.
The probability that $A_1$ doesn't happen, but $A_2$ does, and the rest don't is 
$(1-p_1)p_2(1-p_3)\cdots (1-p_{20})$. 
And so on, $20$ terms. The events described are pairwise disjoint, so the required probability is the sum. 
One can make a simpler expression for this sum by letting $Q=(1-p_1)(1-p_2)\cdots (1-p_{20})$. If none of the $p_i$ is $1$, then the probability that exactly one of the $A_i$ happens is 
$$Q\left(\frac{p_1}{1-p_1}+\frac{p_2}{1-p_2}+\cdots +\frac{p_{20}}{1-p_{20}}\right).$$
