What is the probability that an unfair coin's head appears less than 50 after 100 tosses? I met a question about probability, it seems easy but I got stuck. The question is:
Suppose there is an unfair coin, the HEAD probability is $p=0.7$.
(Q1) If we toss the coin for 100 times, what is the expectation and the variance of this experiment?
(Q2) Answer with reason whether or not the probability is higher than $1/10$
that the number of HEAD appear times is less than $50$ as we toss the coin for $100$ times.
Q1 is easy, I know expectation is $n*p=70$ and variance is $n*p*(1-p)=21$. But for Q2 I have no idea. 
At first I think it looks like... a sample distribution of sample mean used in statistics but... I don't know whether (or how) it will obey a normal distribution. Then I also try to calculate the sum of $P(H=0)+P(H=1)+...+P(H=50)$, but the work is huge, even I use an approximation of Passion distribution...
So could you share some of your thought? Thank you!
 A: We can show that the answer to Q2 is "No" even without appealing to the Central Limit Theorem.
Let's say $H$ is the total number of heads. By the Chebyshev inequality (see below),
$$P(|H-70| \ge 21) \le \frac{21}{21^2} \approx 0.048$$
But 
$$P(|H-70| \ge 21) = P(H \le 49) + P(H \ge 91)$$
so 
$$P(H \le 49) \le P(|H-70| \ge 21) \le 0.048$$

Chebyshev's inequality: If $X$ is a random variable with finite mean $\mu$ and variance $\sigma^2$, then for any value $k>0$,
$$P(|X-\mu| \ge k) \le \frac{\sigma^2}{k^2}$$
A: Since $H$ is a sum of Bernoulli random variables, then $H$ is modeled as Binomial with $N = 100$ trials and $p=0.7$ probability of success. Indeed, it is exhaustive to compute the desired probability which is $$P(H < 50) = P(H=0) + \ldots P(H=49)$$
Instead what you could do is notice that $N =100$ is large enough, and hence we could approximate the Binomial with a Gaussian distribution, so 
$$H \sim N(Np,Np(1-p)) = N(70,21)$$
So
$$\Pr(H<50) \simeq \Pr(\underbrace{\frac{H-70}{\sqrt{21}}}_{Z} < \frac{50-70}{\sqrt{21}}) \simeq \Pr(Z < -4.36)$$
where $Z$ is a centered Gaussian. According to the z-table, the above probability is way less than $\frac{1}{10}$. 
A: Another way to do this is:
$$\sum_{n=0}^{49}\dbinom{100}{n}(0.7)^n(0.3)^{100-n} = 1-\dbinom{100}{50}(0.7)^{50}(0.3)^{50}{{_2}F_1\left(1,-50;51;-\dfrac{7}{3}\right)}\approx 10^{-5}$$
