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 to has to be a 1 AND etc.
(1/2) * (1/2^2) * (1/2^3) * ... * (1/2^10) = (1/2^55) = 2.776*10^-17 was my answer. -Zoe
