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| − | I agree, to maybe make it abit more clear, the same problem we're trying to solve would be akin to flipping a coin 365 times and finding the probability of getting at least one head. obviously the only time you get no heads is if you have all tails, and that will only | + | I agree, to maybe make it abit more clear, the same problem we're trying to solve would be akin to flipping a coin 365 times and finding the probability of getting at least one head. obviously the only time you get no heads is if you have all tails, and that will only be one out of the 2^365 outcomes. |
- Suan-Aik Yeo | - Suan-Aik Yeo | ||
Revision as of 15:00, 16 September 2008
There are two different methods here.... should it be times 365 or raised to the power of 365. And how do you know??? THANKS
// I may be wrong, but I think it should be ^365 instead of *365. If you think about it similarly to a flip of a coin every day for a full year instead of on or off, you would get the number of possible outcomes to be 2^365, either H or T each day. -Evan Clinton
// I agree, to maybe make it abit more clear, the same problem we're trying to solve would be akin to flipping a coin 365 times and finding the probability of getting at least one head. obviously the only time you get no heads is if you have all tails, and that will only be one out of the 2^365 outcomes. - Suan-Aik Yeo
