Probability of being poisoned 
You are playing a game in which you have $100$ jellybeans, $10$ of them are poisonous (You eat one, you die). Now you have to pick $10$ at random to eat.
Question: What is the probability of dying?

How I tried to solve it:
Each jellybean has a $\frac{1}{10}$ chance of being poisonous. Since you need to take $10$ of them, I multiply it by $10$ which gave $1$ (Guaranteed death).
How other people tried to solve it:
Each jellybean is picked out separately. The first jellybean has a $\frac{10}{100}$ chance of being poisonous, the second -- $\frac{10}{99}$, the third -- $\frac{10}{98}$ and so on.. which gives a sum of roughly $\sim 1.04$ (More than guaranteed death!)
Both these results make no sense since there are obviously multiple possibilities were you survive since there are $90$ jellybeans to pick out of.
Can someone explain this to me?

So you live if you do not choose a deadly jellybean :)
And we die if we select at least one deadly bean, so I think it goes as follows
$$P(\text{Die}) = 1-P(\text{Live}) = 1-\frac{\binom{10}{0}\binom{90}{10}}{\binom{100}{10}}=0.6695237889.$$
In this case, we literally have good beans and bad beans, and we select without replacement. Then the number of bad beans selected follows a hypergeometric distribution. So for completeness, there are $\binom{10}{0}$ ways to choose a zero bad beans, $\binom{90}{10}$ ways to choose 10 good beans, and finally $\binom{100}{10}$ ways to choose 10 beans from the total.
Note: $\binom nk = \frac{n!}{k!(n-k)!}$, the binomial coefficient.
