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--[[User:Dakinsey|Dakinsey]] 09:10, 8 October 2008 (UTC) | --[[User:Dakinsey|Dakinsey]] 09:10, 8 October 2008 (UTC) | ||
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| + | ---- | ||
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| + | I got a different result for part 3. Your result only catches the probability 1111111111 will be chosen (1/2 for the first 1, 1/4 for the second...) | ||
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| + | I just did the same as the other two parts, figuring out the probability of 0000000000 and taking 1 minus that number. | ||
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| + | --[[User:Norlow|Norlow]] 15:11, 8 October 2008 (UTC) | ||
Revision as of 11:11, 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)
I got a different result for part 3. Your result only catches the probability 1111111111 will be chosen (1/2 for the first 1, 1/4 for the second...)
I just did the same as the other two parts, figuring out the probability of 0000000000 and taking 1 minus that number.
--Norlow 15:11, 8 October 2008 (UTC)
