if we flip the coin $100$ times, what is$ P(X\leq 10)$?

Question

we have a coin of diameter $d$ and a table of infinite grid of identical squares, each square has side $s$. suppose that $2d = s$. let $X$ denote the total number of times that the coin ends up within a square. if we flip the coin $100$ times, what is $ P(X\leq10)$?

Answer 1

The probability to end up within a square is equal to $\frac 1 4$, so $p\cdot n = 25, p\cdot q\cdot n = 18.75$ and it is OK to use the de Moivre–Laplace theorem. $$P(0\le X\le 10) \approx \Phi(\frac{10-25}{\sqrt{18.75}}) - \Phi(\frac{0-25}{\sqrt{18.75}}) \approx\Phi(-3.46) - \Phi(-5.77)\approx -0,49972 +0,5\approx 0.0002$$

Answer 2

Well, you only have two outcomes for each coin toss: "in a square" and "not in a square". This means you are doing what's called a Bernoulli trial.

By repeating your trial $n$ times (here $n=100$), you are doing a binomial experiment. (You might want to look at this: http://en.wikipedia.org/wiki/Binomial_distribution)

Now if $p$ is the probability of success of your Bernoulli trial ("in a square"), then the probability for having $k$ successes is:

$$P(X=k)={n \choose k}p^k(1-p)^{n-k}$$

The probability for having at most $x$ successes is equal to the sum of probabilities of having $k$ successes, with $k\leq x$:

$$\displaystyle P(X\leq x)=\sum_{k=0}^x P(X=k)=\sum_{k=0}^x {n \choose k}p^k(1-p)^{n-k}$$

Note: The formula comes from the fact that having $k$ successes also means having $(n-k)$ failures (since you only have 2 outcomes). So the probability of having $k$ successes is $p^k(1-p)^{n-k}$. But now you can have these $k$ successes and $(n-k)$ failures in any order among the $n$ trials, this is why you have to add the binomial term ${n \choose k}$

Answer 3

HINT: Say that $p$ is the probability that a coin lands in a square. Then your problem is equivalent to the following:

I toss a coin $100$ times. The coin is biassed and comes up heads with probability $p$. What is the probability that I get heads at most $10$ times?

You might want to look at the binomial distribution.