(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 | ||
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| + | --[[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) | ||
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| + | I have to agree with Dakinsey here, since the question is asking for the probability of the string containing NO 0's, there is only one string (1111111111) that satisfies that condition, so finding it's probability is what is needed. A note about the math involved, since you are taking (1/2)^1 * (1/2)^2 * ... (1/2)^10, you can use a rule of powers and "simplfy" it to (1/2)^(1+2+3+...+10). --[[User:Mnoah|mnoah]] 16:31, 8 October 2008 (UTC) | ||
Latest revision as of 11:31, 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)
I have to agree with Dakinsey here, since the question is asking for the probability of the string containing NO 0's, there is only one string (1111111111) that satisfies that condition, so finding it's probability is what is needed. A note about the math involved, since you are taking (1/2)^1 * (1/2)^2 * ... (1/2)^10, you can use a rule of powers and "simplfy" it to (1/2)^(1+2+3+...+10). --mnoah 16:31, 8 October 2008 (UTC)
