(New page: The first part is pretty straight forward I think. There is only 1 way that all is 1, out of 2^10 ways. But the problem comes at (b), anyone has any idea how to do this?) |
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But the problem comes at (b), anyone has any idea how to do this? | But the problem comes at (b), anyone has any idea how to do this? | ||
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| + | Part b) is just like part a, but the probability of getting a 1 is .6 instead of 1/2. This means that the probability of not having any 0s is the same as having all 1s. So it's just (.6)^10 | ||
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| + | Part c) is similar except that the probability changes for each consecutive 1. The probability of the first digit being a one is 1/(2^i) with i=1. The second is 1/(2^i)=1/4. So you must multiply all of them together like this: | ||
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| + | [1/2] * [1/(2^2)] * [1/(2^3)] * ... * [1/(2^10)] = 2.78e-17 | ||
| + | --[[User:Dakinsey|Dakinsey]] 09:10, 8 October 2008 (UTC) | ||
Revision as of 05:10, 8 October 2008
The first part is pretty straight forward I think. There is only 1 way that all is 1, out of 2^10 ways.
But the problem comes at (b), anyone has any idea how to do this?
Part b) is just like part a, but the probability of getting a 1 is .6 instead of 1/2. This means that the probability of not having any 0s is the same as having all 1s. So it's just (.6)^10
Part c) is similar except that the probability changes for each consecutive 1. The probability of the first digit being a one is 1/(2^i) with i=1. The second is 1/(2^i)=1/4. So you must multiply all of them together like this:
[1/2] * [1/(2^2)] * [1/(2^3)] * ... * [1/(2^10)] = 2.78e-17 --Dakinsey 09:10, 8 October 2008 (UTC)
