Distribution of results of a biased coin the mass distribution of a coin is so that the chance of getting a head s only 40% .The coin 
is tossed a 100 times
what is the chance 
-to get more than (inclusive ) 50 heads ?
-to get more than (exclusive) 50 tails?
to get more than (exclusive) 60 tails and more than (exclusive ) 20 heads ?
if you do the experiment 3times,what is the chance of getting every time less than (inclusive ) 30 heads ?
 A: To get you started, I will work the first one.
Using the Binomial Distribution, we get
$$
\sum_{k=50}^{100}\binom{100}{k}\;.4^k\;.6^{100-k} = 0.027099197757009005051
$$
If we approximate, using the Normal Distribution, where the mean is $-.2\times100=-20$ and the variance is $.96\times100=96$, the probability of getting at least $0$ ($20/\sqrt{96}$ s.d. above the mean) is $\frac{1}{2}(1-\operatorname{erf}(20/\sqrt{96}/\sqrt{2}))=0.020613416668581847098$.
As Brian Scott points out in the comments, we can get a better approximation with the Normal Distribution by including the whole count around $50$ rather than only half the count.  That is, the probability of $49.5$ heads or more. $49.5\text{ heads}-50.5\text{ tails}=-1$ instead of $0$, which gives $19/\sqrt{96}$ or more s.d. above the mean, with a probability of $\frac{1}{2}(1-\operatorname{erf}(19/\sqrt{96}/\sqrt{2}))=0.026239749779623305681$.
A: All right, I'll try to write this a bit more neatly than in robjohn's answer.  He fully disposed of the exact answer; I'll address the approximation by the normal distribution.
It was Abraham de Moivre in the first half of the 18th century who introduced the bell-shaped curve into probability theory, showing that the density is $\varphi(x) = \text{constant}\cdot e^{-x^2/2}$ and finding the constant numerically.  His friend James Stirling showed that the constant is $1/\sqrt{2\pi}.$  Today, the lower-case $\phi$ is standard notation for this function and capital $\Phi$ is its antiderivative that approaches $0$ at $-\infty$ and $1$ at $\infty$.
When one tosses a biased coin, one gets 1 head with probability $p$, or 0 heads with probability $1-p$.  The expected value of that probability distribution is $0\cdot(1-p)+1\cdot p = p$.  The standard deviation is $\sqrt{p(1-p)}$, and I leave the details to you unless you ask.
When one tosses it 100 times, and the tosses are independent, the expected value is $100p$ and the standard deviation is $\sqrt{100p(1-p)}$.  For $p=0.4$, the expected value is 40 and the standard deviation is $\sqrt{100p(1-p)}\approx 4.898979485566$.
The probability that this is at least 50 is also the probability that it is strictly more than 49.  The "continuity correction" calls for using the number half-way between those when doing the continuous approximation.  That is $49.5$.
So the probability of getting 49 or fewer successes is
$$
\Phi\left(\frac{49.5 - 40}{4.898979485566}\right) \approx 0.97376.
$$
The probability of getting 50 or more successes is 1 minus that number.
The exact answer for $\le49$, if I can believe the software I'm using is about $0.9729008$.
If you use the continuity correction, then even with a sample of size 1 (where it would be silly---and worse, it would be pointless---to use the continuous approximation) the continuous approximation never differs from the exact result by more than about $0.03$.  With six coin tosses, it gives somewhat respectable results.  With 100, it is reasonably accurate.
