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!
Best Answer
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 $k>0$, $$P(|X-\mu| \ge k) \le \frac{\sigma^2}{k^2}$$