Question: **Question = Bold**

What I know = Regular

(a) **If E, G are going, what's the probability A and C are going?** 

-From this, all I know is that there are 7 letters to choose 2 from for those who are going, since it's already given that E and G are going. 

(b) **If A is not going, what's the probability B and C are going?** 

-There are eight letters that one can choose four from, since A is not going. 

(c) **If A, C, D are not going, what's the probability that E, B, F are not going?** 

I'm not quite sure on the logic to get the correct answers. Can someone please explain how to do it, because I honestly don't understand how to approach these problems. Thanks in advance.

Answer: 
Suppose $f(x)=x^2-2$ so roots are $\pm \sqrt2$
now use Newton method $$x\_{n+1}=x\_n-\frac{f(x\_n)}{f'(x\_n)}$$
so you will have 
$$x\_{n+1}=x\_n-\frac{x\_n^2-2}{2x\_n}\\=\frac{\frac{2x\_n^2-x\_n^2+2}{x\_n}}{2}\\=\frac{x\_n+\frac{2}{x\_n}}{2}$$ now take $x\_n \to a\_n$ so 
$$a\_n = \frac {a\_{n-1} + \frac {2}{a\_{n-1}}}{2}$$ and note $a\_n $ tends to $\sqrt 2 ,if \space a\_1>0$ , tends to $-\sqrt 2 ,if \space a\_1<0 $ by iteration.  

$\bf remark:$ there are some equation like this which work with rational numbers and finally get irrational ... for example :$\sum \_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}\\\text{sum of rationals = irrational}$
Because if $a$ is the limit then $a^2=2$, which gives $a=\sqrt2$, which is an irrational number. 

$$a\_{n}-\sqrt2=\frac{(a\_{n-1}-\sqrt2)^2}{2a\_{n-1}}>0$$ and
$$a\_n-a\_{n-1}=\frac{1}{a\_{n-1}}-\frac{a\_{n-1}}{2}=\frac{2-a\_{n-1}^2}{2a\_{n-1}}<0,$$
which gives that our sequence converges.