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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?

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

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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:

$$\Pr(N = n)={4\choose n}0.10^n0.90^{4-n}$$

However, you want the chance that at least one attempt succeeds, which can be more easily found by examining converse: the chance that no attempt succeeds

$$\Pr(N \geq 1)=1-\Pr(N = 0)$$

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