Union of three independent events Let  $ A_1, A_2, A_3 $ be independent events with probabilities $ \frac 12,\frac13,\frac14,$ respectively. How to compute for  $P(A_1\cup A_2 \cup A_3).$

My solution starts from using the probability of their complements, I do not know how to answer this question. Please help.

$P(A_1\cup A_2 \cup A_3) = 1- P(A_1^c\cap A_2^c \cap A_3^c)$
$P(A_1\cup A_2 \cup A_3)= P(A_1\cup A_2) + P(A_3) - P((A_1\cup A_2)\cap A_3 )$
$=P(A_1\cup A_2) + P(A_3) - P(A_1\cup A_2).P(A_3 )$ ..since $(A_1\cup A_2)$ and $(A_3)$  are independent

Similarly,
$P(A_1\cup A_2)=P(A_1)+P(A_2)-P(A_1\cap A_2)=P(A_1)+P(A_2) - P(A_1).P(A_2)$

Now you can just do the calculation. I guess the final answer is 3/4
Use the following fact 


  If $A$ and $B$ are independent events, then $A^{c}$ and $B^{c}$ are independent evetns


Proof: We see that 
$$P(A^c\cap B^c)=P(A^c)+P(B^c)-P(A^c\cup B^c)=P(A^c)+P(B^c)-P((A\cap B)^c)=$$
$$P(A^c)+P(B^c)-(1-P(A\cap B))$$
which since $A$ and $B$ are independent then we have 
$$=P(A^c)+P(B^c)-(1-P(A)P(B))=(1-P(A))+(1-P(B))-(1-P(A)P(B))=$$
$$1-P(A)-P(B)+P(B)P(A)= (1-P(A))(1-P(B))=P(A^c)P(B^c)$$

Thus $A^c$ and $B^c$ are independent