Is there a faster way to get a definite numerical value for the following permutation/ combinaions question? 
The ratio of boys to girls at birth in Singapore is 1.09:1 Then the
proportion of Singaporean families with exactly 6 children who will
have at least 3 boys is
A. 0:696
B. 0:315
C. 0:521
D. 0:455

While i know the answer should be $$\binom{6}{3}\frac{1.09}{2.09}^{3}\frac{1}{2.09}^{3}+\binom{6}{4}\frac{1.09}{2.09}^{4}\frac{1}{2.09}^{2}$$......
Is there any way in which i can come to the 0.696 value without a calculator fast enough?
We are not allowed calculators in exams and the sample paper contains this question.
Thanks a ton in advance.
 A: This is not a strict prove at all and in general situation with high orders can work absolutely wrong, but I think you can reason in this way:
$\binom 63 = 20$
$\binom 64 = 15$
$\binom 65 = 6$
$\binom 66 = 1$
So your final probability expression could be written like this:
$(\frac{1.09^3}{2.09^6})*(20+15*1.09+6*1.09^2+1*1.09^3)=(\frac{1.09^3}{2.09^3})*(\frac{20+15*1.09+6*1.09^2+1*1.09^3}{2.09^3})$
The first fraction $(\frac{1.09^3}{2.09^3})$ i think could be approximately rewritten as $(\frac{1.09^3}{2.09^3})\approx(\frac{1}{2})^3=0.125$
The second $(\frac{20+15*1.09+6*1.09^2+1*1.09^3}{2.09^3})\approx(\frac{42}{8})=5.25$
Thus your result will be $\approx0.125*5.25=0.65625$ which is pretty close to the correct one.
A: You need two more terms in your sum, for five and six boys.  I don't know a faster way to compute the actual value, but it is easy to see that among the choices given the answer must be A.  The chance has to be greater than $0.5$, which eliminates B and D.  If there were the same number of boys and girls, the chance of having at least three boys in six would be higher than $0.521$ because we succeed with a $3-3$ tie.  Having more boys than girls only raises the ratio.  $0.521$ is too small to be believable.  
To be exact, assuming equal numbers of boys and girls, the chance of at least three boys is $\frac 12+\frac 12{6 \choose 3}2^{-6}=\frac 12 + \frac {20}{128}=\frac {21}{32}\approx 0.656$  Having more boys than girls only raises this. 
