Binomial theorem in probability We know according to binomial probability theorem ,
$$P= \binom{n}{r} p^r (1-p)^{n-r} \tag{1}$$
Now If I flip a coin 10 times and want to get the probability for 4 heads then we get according to the binomial theorem:
$$P= \binom{10}{4} \left({\frac{2}{5}}\right)^4\left(1-\frac{2}{5}\right)^{6}$$
Am I right?
If I'm right then what would be for the Newton–Pepys problem.
Why did they use the probability 
$$P= 1-\binom{n}{r} p^r (1-p)^{n-r}$$

Cant we use eq(1)?


Yes you can use the first formula, but it would just take too long. For example "Six fair dice are tossed independently and at least one “6” appears.": You would have to add up the probabilities of 1 to 6:
$$\sum_{k=1}^{6}\binom 6 k(\dfrac{1}{6})^k(\dfrac{5}{6})^{6-k}=0.6651$$
Whereas in formula 2 you only use 1-(Probability of Failure)"... 
PD: What sort of coin are you using in your example?
