A $100$ page book has $200$ printing errors randomly distributed throughout. What is the probability that one of the pages is error free?
Consider $$\sum _{i=1}^{100}x_i = 200$$ where each $x_i$ can take integer values from $0$ to $200$. This gives a total of $\binom{299}{99}$ possible ways in which the errors can be distributed in the book. Is it sufficient to now obtain the result as $$\sum_{i=1}^{99}p(i)$$ where $p(i)$ is the probability that exactly $i$ pages are error free?
Best Answer
Its a possions distribution
$P= \lambda^x * e^(-\lambda)/x!$ take $\lambda$$=200/100=2$ and $x=0$ you will get the solution around 13%.