Probability that a fair coin is tossed more than 100 times until 40 tails are observed What is the probability that a fair coin is tossed more than 100 times until 40 tails are observed? 
I know when a fair coin is tossed 100 times, the probability of 40 tails are observed, but I wonder what concept should I use for this question.
 A: Well, there are two possible event types to consider: it takes $100$ tosses at most for $40$ tails to be observed, or it takes more than $100$ tosses. The latter, of course, is that for which you seek the probability. This gives us two straightforward ways to approach the problem.


*

*Breaking down the first type into cases: the $40$th toss is the $40$th tails toss, the $41$st toss is the $40$th tails toss, ..., the $99$th toss is the $40$th tails toss, the $100$th toss is the $40$th tails toss. Adding the respective probabilities of these cases, we find the probability of the first type of event, whose complement is the desired probability.

*Breaking down the second type into cases: there are no tails tosses in $100$ tosses, there is $1$ tails toss in $100$ tosses, ..., there are $38$ tails tosses in $100$ tosses, there are $39$ tails tosses in $100$ tosses. Adding the respective probabilities of these cases gives us the desired probability.

A: This is tantamount to heads falling 61 times or more out of 100 tosses.
This probability is calculated by using the binomial probability distribution.
The probability of heads falling 61 times is 
$${100\choose61}\cdot0.5^{61}\cdot 0.5^{39} = {100\choose61}\cdot 0.5^{100}$$
The probability of heads falling 62 times is 
$${100\choose62})\cdot 0.5^{62}\cdot 0.5^{38} = {100\choose62}\cdot0.5^{100}$$
................................... and so on. 
The final term is $${100\choose 100}\cdot0.5^{100}$$
So, your probability is the sum of all these probabilities:
$$0.5^{100}\Bigg[{100\choose61} + {100\choose62} +.. + {100\choose100}\Bigg]$$
You can of course arrive at the same answer by saying that it is tantamount to the probability of tails falling 39 times or less. The solution is identical:
$$0.5^{100}\Bigg[{100\choose0} + {100\choose1} + {100\choose2}+...+ {100\choose39}\Bigg] = 0.5^{100}\Bigg[{100\choose61} + {100\choose62} +.. + {100\choose100}\Bigg]$$
You need to write a program on your calculator to add all these numbers fast if your calculator allows it, or use a program/website. It can be done semi-manually much faster where you should use a key for combinations $_nC_r$ and a memory key on your calculator. You should discard low numbers, such as C(100,80) and lower, or you can discard even C(100,75) or lower. They are just a waste of time unless you need a lot of precision.
The answer will be ≈0.0176 or ≈1.76% (done on a calculator). 
You can also use the de Moivre-Laplace central limit theorem for a faster calculation. All I did is showed an elementary solution here. Probably you were asked to apply that theorem because the numbers in your problem are pretty large. You can just plug the data into the formula and use a program to calculate the necessary integral. I’m not writing the formula (Please, look it up. It’s called de Moivre-Laplace limit theorem). There's also an option of using the table with $Φ(x)$. If it's statistics you are probably expected just to apply the $Φ(x)$ from the table ($Φ(x)=0.9821$. The continuity correction will be needed!) making the task look as simple as possible. It's less precise though anyway. Your probability is 
$1-Φ(x)=0.018.\;$ You have nice round figures in your problem ($X=60.5$ with the continuity correction):
$$\frac{X-100\cdot0.5}{\sqrt{100\cdot0.5\cdot0.5}}≥2.1$$ Hence, $Φ(x)$=0.9821
Sorry. I’m still a bit bad with LaTeX. I need to get accustomed to it. So I gave you a “dumbed down” solution but it works perfectly fine. I know I should probably have given you the solution with the formula. But it should not be a problem to apply the integral, although it may look intimidating. 
Hope my explanations were not confusing.
A: The concept you require may be the negative binomial distribution for the distribution of the number of times a fair coin is tossed until $40$ tails are observed. This would give $$\sum_{k=101}^{\infty} {k-1 \choose k-40} \frac{1}{2^{k}}$$ though you might find this easier as a finite sum: $$ 1-\sum_{k=40}^{100} {k-1 \choose k-40} \frac{1}{2^{k}}$$
You can also find the answer to your particular question using the ordinary binomial distribution by rewriting it as 

What is the probability that a fair coin tossed $100$ times leads to fewer than $40$ tails being observed?

making the answer $$\sum_{n=0}^{39} {100 \choose n} \frac{1}{2^{100}}$$
So in R the following pieces of code each produce the same answer:
pnbinom(60, 40, 1/2, lower.tail = FALSE)

1 - sum(choose(39:99, 0:60) / 2^(40:100))

sum(choose(100, 0:39) / 2^100)

pbinom(39, 100, 1/2)

If you wanted to use the normal approximation with a continuity correction, then as Ken Draco says in his answer, $\Phi\left(\dfrac{40-0.5-100\times \frac12}{\sqrt{100\times \frac12\times \frac12}}\right)=\Phi(-2.1)$ would be reasonably close 
