A Question Of Percentages

If I have 4 chances and each chance has a 10% success rate, what is the overall percent chance that 1 chance will succeed? For example: A guy plays a roleplaying game. He has 4 peices of equipment where each peice has a 10% chance rocks will fall from the sky. The percentages don't stack. Meaning they don't add up to give a 40% chance. Each one is worked out independantley. So what would be the percent chance 1 will succeed out of all the 4?

• What are your thoughts on the problem? – Deepak Jul 18 '14 at 3:01
• Are the chances independent? – copper.hat Jul 18 '14 at 3:10
• Maybe you should look at the chance of complete failure? – copper.hat Jul 18 '14 at 3:11
• Some advice on asking good questions... – user147263 Jul 18 '14 at 3:18
• What does "overall chance that I succeed" mean to you: succeed in all attempts, or in at least one attempt? – Graham Kemp Jul 18 '14 at 3:21

Assuming the chances of success on each attempt are mutually independent (and equal), then the number of successes in the trial has a binomial distribution. $N\sim\mathcal{Bin}(4, 0.10)$
So the chance that exactly $n$ attempts succeed is determined by:
$$\Pr(N = n)={4\choose n}0.10^n0.90^{4-n}$$
$$\Pr(N \geq 1)=1-\Pr(N = 0)$$