The other answers are great, but with a slightly different approach from your attempt (I presume you edited in your attempts after some of the answers are posted). I'd like to point out some important things in your post:
"And" vs. "Or"
One fundamental flaw in both approaches is that you are adding probabilities instead of multiplying them.
When you are looking for either of a set of independent events, you can add their probabilities. For example, suppose that you roll a die. The probability for a 1 is 1/6, for a 2 is 1/6, etc. The probability that you roll an even number is the probability of either 2 or 4 or 6 so you can add the probability for these events: 1/6 + 1/6 + 1/6 = 1/2. This probably corresponds to you intuition: half of the possible outcomes are even.
Now suppose you throw a single die three times (or three dice at once). What is the probability that all rolls are even? We just saw that the probability for any of them to be even is 1/2. But if you try 1/2 + 1/2 + 1/2 = 3/2 > 1. Clearly this cannot be right. Indeed, what you are after is a combination of events: you want the first roll to be even and the second to be even and the third. What you need to do is multiply the probabilities: 1/2 * 1/2 * 1/2 = 1/8.
You can check this by writing out all possible outcomes: let
E denote even and
O denote odd, then all possibilities for three dice rolls are
EEE - 1 with only evens
EEO, EOE, OEE, - 3 with 2 evens, 1 odd
EOO, OEO, EOO - 3 with 1 even, 2 odds
OOO - 1 with only odds
Do this systematically, and you will find 8 possible outcomes, of which only one is
EEE -- 1/8.
With vs without replacement
There is a difference whether you pick with replacement or without replacement. Suppose you pick a marble from an urn with 5 white and 5 black marbles. Clearly the probability of drawing a white marble is 1/2. Now what is the probability that the next one will be white again? Well, this depends on what you do with the first one! If you put it back in the urn after you draw it, the probability that the second one is white is 1/2 again! But if you don't put it back, then you only have 4 out of 9 whites left, so the probability is 4/9 -- less than 1/2!
In your example with the jelly beans, you are eating them, so it is an experiment without replacement. Once you had a non-poisonous one (with probability 90/100), the chances that the next one also won't kill you are not 90/100 anymore, they are 89/99.
All possibilities or complement rule?
There are now two ways to arrive at your answer:
- Enumerate all the possibilities. You are asked for the probability of dying. There are a number of ways this can happen: either the first bean kills you, or (it doesn't and the second one kills you), or (the first two don't kill you but the third one does), or ....
Note how I emphasized "or" and "and"? Replace them by
x as explained above, and you will get a long calculation:
P(first kills you)
+ P(first doesn't kill you) * P(second kills you)
+ P(first doesn't kill you) * P(second doesn't kill you) * P(third kills you)
Plugging in the numbers, keeping in mind you are working without replacement, you get
+ 90/100 * 10/99
+ 90/100 * 89/99 * 10/98
Use the complement rule. Sometimes it is easier to calculate the probability of the event you are not interested in. You know that either the 10 beans kill you or they don't, so
P(you die) + P(you live) = 1
P(you die) = 1 - P(you live).
We have seen that
P(you die) is a long calculation, but
P(you live) is easier:
P(you live) = P(the first one doesn't kill you)
* P(the second one doesn't kill you)
* P(the third one doesn't kill you)
Note that I used multiplication, convince yourself why (is there an implicit "and")?
Now I suggest you complete both calculations (1) and (2) and check that they give the same answer. It will be a lot of work but it will give you confidence that the basic rules I laid out work, and that there is not one right way.