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Does anyone understand part C. It's confusing me. --[[User:Krwade|Krwade]] 20:48, 24 February 2009 (UTC) | Does anyone understand part C. It's confusing me. --[[User:Krwade|Krwade]] 20:48, 24 February 2009 (UTC) | ||
| − | Here's how I did part c. Probability of getting no 0's = probability of getting all 1's. For each digit i, probability of getting a 1 is 1/2^i. Multiply all the probabilities [because digit #1 has to be a 1 AND digit #2 has to be a 1 AND etc.]. | + | Here's how I did part c. Probability of getting no 0's = probability of getting all 1's. For each digit i, probability of getting a 1 is 1/2^i. Multiply all the probabilities [because digit #1 has to be a 1 AND digit #2 has to be a 1 AND etc.]. So, |
(1/2) * (1/2^2) * (1/2^3) * ... * (1/2^10) = (1/2^55) = 2.776*10^-17 was my answer. -Zoe | (1/2) * (1/2^2) * (1/2^3) * ... * (1/2^10) = (1/2^55) = 2.776*10^-17 was my answer. -Zoe | ||
Latest revision as of 16:23, 25 February 2009
I think for part b.) we use Bernoilli trials, so the solution would be found by (10 choose 0)*(.6^10)*(.4^0)
Does anyone understand part C. It's confusing me. --Krwade 20:48, 24 February 2009 (UTC)
Here's how I did part c. Probability of getting no 0's = probability of getting all 1's. For each digit i, probability of getting a 1 is 1/2^i. Multiply all the probabilities [because digit #1 has to be a 1 AND digit #2 has to be a 1 AND etc.]. So,
(1/2) * (1/2^2) * (1/2^3) * ... * (1/2^10) = (1/2^55) = 2.776*10^-17 was my answer. -Zoe
