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: 
Yep, the author knew the answer beforehand, I would say much longer before.

Because this is just an instance of the Babylonian method of computing square roots, later discovered to be a particular case of Newton's iterative method for the resolution of nonlinear equations.

So with closed eyes, this sequence converges to $\sqrt2$. You can easily verify it be assuming convergence, so that $a\_{n-1}$ and $a\_n$ become indiscernible, and

$$a=\frac{a+\dfrac2a}{2}$$ or $$a^2=2.$$ As the initial value is $1$, all terms are positive and convergence is to the positive root (if there is convergence, though).

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There is a simple way to explain that method, also known as Heron's formula.

Let $s$ be the number of which you want to extract the root, and let $a$ be an approximation by default. Then $$a<\sqrt s\implies a':=\frac sa>\sqrt s$$ so that $\dfrac sa$ is another approximation, by excess. Now if we take the arithmetic mean, we get a new approximation which is closer than the worse of the two,

$$a''=\frac{a+a'}2=\frac{a+\dfrac sa}{2}.$$

As can be shown, when you are close to the root, the sequence converges extremely rapidly.

For example,

$$a=\color{green}{1.41}\implies a'=\color{green}{1.41}84397163121\cdots\implies a''=\color{green}{1.41421}9858156\cdots$$ while the true value is
$$\sqrt2=1.4142135623731\cdots$$

The next iteration gives $11$ exact digits.

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A note on convergence, for the skepticals (the method has been in use for at least two millenia).

Let $$x\_n:=\frac{a\_n}{\sqrt s}.$$

We have

$$\frac{x\_{n+1}-1}{x\_{n+1}+1}=\frac{x\_n+\dfrac1{x\_n}-2}{x\_n+\dfrac1{x\_n}+2}=\frac{(x\_n-1)^2}{(x\_n+1)^2}$$
and by induction

$$\frac{x\_n-1}{x\_n+1}=\left(\frac{x\_0-1}{x\_0+1}\right)^{2^n}.$$

This is an exact formula for the $n^{th}$ iterate, which proves convergence from any $x\_0$ such that

$$\left|\frac{x\_0-1}{x\_0+1}\right|<1.$$

This holds for all positive $x\_0$.

In passing, this also proves quadratic convergence, i.e. the relative error is squared on every iteration.
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.