Probability of being poisoned

Both probablyme and carmichael561 have given a good approach to the problem, but I thought I'd point out why the solutions given by you and your classmates (?) are erroneous.

The problem common to both approaches is that they neglect the probability that you die from earlier jelly beans. You take the first jelly bean, and you have a $1/10$ probability of dying; that is all right. But although your classmates are almost right (and you are not) that the second jelly bean has a $10/99$ chance of killing you, that is only true if the first jelly bean didn't already kill you.

In other words, the probability that you are killed by the first jelly bean is $1/10$, and the probability that you survive to eat a second jelly bean and it kills you is $9/10 \times 10/99 = 1/11$. Each succeeding jelly bean does, in truth, have a higher probability of killing you if you survive to eat it, but the decreasing probability that you do in fact survive conspires to make its overall effect smaller, so the numbers do not add up to anywhere near $1$, complete certainty.

It is possible to continue along in a similar vein: You can add $1/10+1/11 = 21/110$ to obtain the probability that the first two jelly beans kill you; the remainder, $89/110$, is the probability that you survive to eat the third jelly bean, which kills you with probability $10/98$. Your probability of surviving the first two jelly beans only to be killed by the third is then $89/110 \times 10/98 = 89/1078$. You would then have to add up $1/10+1/11+89/1078$ to find the probability that the first three jelly beans kill you, etc.

I think you can see that the solutions provided by the other answerers are much more straightforward; this way of approaching the problem by considering its inverse is a common tactic when there are multiple ways to satisfy the conditions of the problem, but only one way to violate them.

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.

It's easier to find the complementary probability; i.e. the probability that all $10$ jellybeans are safe. There are ${90\choose 10}$ ways to choose $10$ safe jellybeans, and ${100\choose 10}$ to choose any $10$ jellybeans, so the probability that you survive is $$ \frac{{90\choose 10}}{{100\choose 10}}=\frac{90\cdot 89\cdots 81}{100\cdot 99\cdots 91}$$

The probability that you die is therefore $$ 1-\frac{90\cdot 89\cdots 81}{100\cdot 99\cdots 91}\approx 0.67$$