Since when is 9/10 = 92%? This is probably a more basic question than this site is used to, probably because I'm only 13, and as such I'd appreciate if you gave a more basic and simple explanation than the norm for this site.
I was reading a BBC News article this morning and I found myself questioning the three diagrams, in the 'Why are two tests better than one?' section.
You do a first test and obtain nine heads and one tail... The probability that the coin is fair given this outcome is about 8%, [and the probability] that it is biased, about 92%.
You do a second test, and this time you throw eight heads and two tails. Now the probability for a fair coin is about 16%, for a biased coin about 84%.
So the naive thought might be that you haven't gained any certainty from this second test.
But if you think about it differently, what you've really done is throw the coin 20 times and get 17 heads and three tails." This means there's a probability of 98.5% that the coin is biased.
I have three questions, all roughly along the same lines: Please see below paragraph, these questions are no longer 'active'
 - Since when is 9/10 heads a 92% probability, and why?
 - Why is 8/10 an 84% probability, I always thought that 8/10 = 80%
 - And finally, 17/20 is 85%, except when it's 98.5%. Why?
After some help from the comments, I now realise that it's not talking about the probability of getting a head (or a tail) but the probability of the coin being biased. Can someone explain (preferably in layman's terms) how the article gets to 92%, 84% and 98.5% respectively?
Thanks.
 A: Both you and the BBC news author suffer from a misunderstanding.
If a coin produces nine heads and one tail in ten tossings, it is reasonable to extrapolate and assume that even for $n\gg 10$ we may expect about $\frac9{10}n$ heads and $\frac 1{10}n$ tails. That is: By observing a random experiment (whatever that is) often enough we estimate an underlying probability that allows us to predict future repetitions of this experiment. This is where you get your $\frac 9{10}$ or $90\,\%$ from.
If a fair coin (i.e. one where heads and tails occur with probability $\frac12$ each) is tossed $n$ times, then the probability of exactly $k$ heads in this sequence is $\frac{n\choose k}{2^n}$. Expecially, with $n=10$ a sequence with exactly $9$ heads will occur with a probability of only $\approx0.0098%$. Actually, even an outcome of exactly five heads is not very common as it occurs with only probability $\approx 0.246$. So we should rather ask: What is the probability of an overdose of either heads or tails as extreme as obserevd or even more extreme? That is: What is the probability that nine or more heads or nine or more tails occur? This probability turns out to be $\approx0.021$. That is: in about $98\,\%$ of all experiments where you perform ten tosses with a coin known to be fair, you will observe that one of the two outcomes is extremely underrepresented (occurs at most once). This does not say, however, that the coin is not fair (after all I specified that the coin is known to be fair). This is also like the second part of the article says, that among sufficiently many nurses, even if they are "fair coins", a certain small percentage will trigger a test that is doomed to trigger a small percentage.
What the BBC article (or rather Colmez) claims about the coins, however, is that the experiment shows that the probability that the coin is biased is $92\,\%$. This is something that cannot be concluded in this extend in any way.
Assume you have a big bag of $N$ coins. You know that one of these coins is biased - it produces heads with probability $p>\frac12$ instead of $\frac12$.
You draw a coin at random. What is the probability that it is faked? Obviously, without further tests, it is $P(B)=\frac1N$ (where $B$ denotes the event that the coin is biased, so $\overline B$ is the event that the coin is fair).
Now you perform our test, ten tosses, and event $A$ "nine heads" occurs. What do we learn from that?
For a fair coin, $A$ occurs with $P(A|\overline B)={10\choose 9}\cdot 2^{-10}$, for a fake coin, $A$ occurs with $P(A|B)={10\choose 9}\cdot p^9(1-p)$. As we don't know which coin we used, we know beforehand only that
$$P(A)=P(\overline B)P(A|\overline B)+P(B)P(A|B)=\left(1-\frac1N\right){10\choose 9}\cdot 2^{-10}+\frac1N{10\choose 9}\cdot p^9(1-p)$$
Now Bayes theory comes in and says that
$$ P(B|A) =\frac{P(B\land A)}{P(A)} = \frac{\frac1N{10\choose 9}\cdot p^9(1-p)}{\left(1-\frac1N\right){10\choose 9}\cdot 2^{-10}+\frac1N{10\choose 9}\cdot p^9(1-p)} = \frac{ p^9(1-p)}{\left(N-1\right) 2^{-10}+ p^9(1-p)}.$$
Well, obviously this expression depends on both $N$ and $p$ and can hardly be equated to $92\,\%$ just like this. For example if we had $p<\frac12$, then the observation of $A$ would even lower the probability of the coin being faked. Or if the fake coin has heads on both sides ($p=1$) it is even impossible that the coin is the faked one.
Even in an "optimal" situation where the coin has $p=\frac9{10}$ (so that nine heads, one tails would be the most likely outcome for that coin), we find
$$ P(B|A) =\frac{387420489}{9765625N + 377654864}. $$
For big $N$ this is 
$$P(B|A)\approx39.7\cdot \frac1N,$$
so this test can hardly tell us that the coin we have in our hands is the biased one with probability $92\,\%$ if $N$ is sufficiently big! With $N=1000$ coins in our bag, we could raise our confidence from one permille to about four percent, way less than $92\,\%$. In fact, if you only have four coins ($N=4$) and one of them is biased such that it makes the test most predictive (i.e. the one biased coin has $p=\frac9{10}$), then  the test (if it results in nine heads as outcome!) allows you so say that with probability $93,\%$ the coin in your hand is the heavily biased one (as opposed to that you could only make the same claim with $25\,\%$ confidence before the test).
Nevertheless, the main point is correct: If you perform several independent(!) tests with low reliability, then collectively these can give high reliability. A single photon hititng your retina cannot tell you anything and yet with a few gazillion photons you recognize exactly what goes on in your environment. 
A: the definition you use for the probabilities is $\frac{N_{cases}}{N_{total-cases}} $, which is a definition that applies in certain circumstances. But the way probablities are computed in this problem are different. They involve a more complex problem that has a more complex response. The question is, what is the probability of getting 9 heads an one tail, if the coin is not biased, and what is the probablity that a casee like that is the result of a bias. They involve a more advanced study of probablility theory beyond the definition that you use.
