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Hey, It's me again!

Before I go on with this question, I want to thank you Dalton (and Cia), you saved me from going insane over that question

But, unfortunately, the hard sums of the universe strike again! (Oh noes) and its on the same topic as last time, chance.

Let's say I'm playing an unusual game where I have 3 sweets, and I want to eat as much of them as possible.

There is a 90% chance that I will be able to eat the first sweet. If I don't eat it, it's game over and I can't eat any more sweets anymore

If I do eat it, I have an 80% chance to eat the next sweet. If I don't eat it, it's game over, like above.

If I eat the second sweet, I have a 70% chance of eating the final sweet.

And there will be no more sweets left.

How do I find out my chances to

a) Eat at least 2 sweets

b) Eat all 3 sweets

Thanks in advance,

~Hammy

Well ideally, for health reasons, you should eat NONE of those sweets. XD

But in any case, since I'm short on time, I'll have to hope a professional math wiz can come and help.  

I don't know if I'm allowed to do this, but I'll help anyway.
Remember that these are independent events, and that one result doesn't affect the value of the others.
A) 9/10 x 8/10 = 72/100
B) 9/10 x 8/10 x 7/10= 504/1000
(Multiply the numerators and denominators)
Sorry if I'm bad at writing these things XD it would be 10000x easier if I had a pencil/paper.

as cykrion 2.0 said, you simply multiply the odds of each event happening, because the events are independent. I can explain why we multiply the odds, if further explanation is needed, but Cykrion's answers are correct, if that is all that is needed.

His answers reduce down to 18/25 and 63/125, in their simplest forms.

Yay!  Math wizards to the rescue! XD